The time taken for the stone to hit the ground can be calculated using vertical motion equations, and the stone's speed at impact can be determined using the vertical velocity equation.
Time taken for the stone to hit the ground can be calculated using the vertical motion equation: t = sqrt(2h/g). Substituting the values, t = √(2×45/9.81) ≈ 3.0 seconds.
Stone's speed at impact can be calculated using the vertical velocity equation: vf = u + gt, where vf is the final velocity (0 m/s at impact), u is the initial vertical speed, and g is the acceleration due to gravity.
With the given information, the stone's speed at impact is approximately 29.4 m/s.
What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant velocity, if the coefficient of friction between the plane and the block is μ? Group of answer choices
a. mg μ cosθ
b. mg (μ cosθ + sinθ)
c. mg (μ sinθ + cosθ)
d. mg (μ cosθ + μ sinθ)
e. mg (μ cosθ + m sinθ)
The force needed to push a block up an incline at a constant velocity is the sum of the gravitational force parallel to the incline and the frictional force, which is mg sin(θ) + μmg cos(θ). So the correct option is e.
Explanation:The force required to push a block of mass m up an inclined plane at an angle θ with the horizon at a constant velocity, given the coefficient of friction between the plane and the block is μ, can be determined by analyzing the forces acting on the block. We need to consider both the component of the block's weight parallel to the incline and the frictional force.
Since the block is moving at a constant velocity, the net force along the incline must be zero. This implies that the applied force must counteract the combined forces of gravity pulling the block down the incline and the frictional force opposing the motion. The gravitational component parallel to the incline is mg sin(θ) and the frictional force is μmg cos(θ). Therefore, the required force F is:
F = mg sin(θ) + μmg cos(θ)
This corresponds to answer choice e.
We've seen that bees develop a positive charge as they fly through the air. When a bee lands on a flower, charge is transferred, and an opposite charge is induced in the earth below the flower. The flower and the ground together make a capacitor; a typical value is 0.60 pF. If a flower is charged to 30 V relative to the ground, a bee can reliably detect the added charge and then avoids the flower in favor of flowers that have not been recently visited Approximately how much charge must a bee transfer to the flower to create a 30 V potential difference?
To solve the problem it is necessary to apply the concept of Load on capacitors. The charge Q on the plates is proportional to the potential difference V across the two plates.
It can be mathematically defined as:
Q= CV
Where,
C = Capacitance
V = Voltage
Our values are given as,
[tex]C = 0.60pF\\V = 30V[/tex]
Substituting values in the above formula, we get
[tex]Q=CV\\Q = 0.6*30\\Q = 18pC[/tex]
Where
[tex]1pC = 10^{-12}Coulomb[/tex]
Therefore the charge must be 18pC to create a 30V Poential difference.
A garden hose with a small puncture is stretched horizontally along the ground. The hose is attached to an open water faucet at one end and a closed nozzle at the other end. As you might suspect, water pressure builds up and water squirts vertically out of the puncture to a height of 0.73 m. Determine the pressure inside the hose. (Enter your answer to the nearest 1000 Pa.) 7161.3 If you use the subscripts "1" and "2" respectively to represent a point in the hose before the leak and the site of the leak, how does h compare to h2? How does v1 compare to v2? What is the value of the pressure at the site of the puncture?
Answer:
The pressure inside the hose 7000 Pa to the nearest 1000 Pa.
[tex]h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}[/tex]
[tex]V_{1} =V_{2}[/tex]
The pressure at the site of the puncture is [tex]P₂ =7161.3 Pa[/tex]
Explanation:
According to Poiseuille's law, [tex]P_{1} - P_{2} = \frac{128μLQ}{πD^{4} }[/tex]
Where [tex]P_{1}[/tex] is the pressure at a point [tex]1[/tex] before the leak, [tex]P_{2}[/tex] is the pressure at the point of the leak [tex]2[/tex], μ = dynamic viscosity, L = the distance between points [tex]1[/tex] and [tex]2[/tex], Q = flow rate, D = the diameter of the garden hose.
Also, from the equation [tex]P =ρgh[/tex], the equations [tex]h₁ = \frac{P₁} {ρg}[/tex] and [tex]h₂ = \frac{P₂} {ρg}[/tex] can be derived.
Combining Poseuille's law with the above, we get [tex]h₁ - ρgh₂ = \frac{128μLQ}{πD^{4} }[/tex]
[tex]h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}[/tex]
[tex]V =\frac{Q}{A}[/tex]
Since the hose has a uniform diameter, the nozzle at the end is closed and neither point [tex]1[/tex] nor [tex]2[/tex] lie after the puncture,
[tex]V_{1} =V_{2}[/tex]
The pressure at the site of the puncture [tex]P₂ =ρgh₂[/tex]
[tex]P₂ =1kgm⁻³ ×9.81ms⁻¹×0.73m[/tex]
[tex]P₂ =7161.3 Pa[/tex]
A student stands on the edge of a merry-go-round which rotates five times a minute and has a radius of two meters one evening as the sun is setting. The student produces a shadow on the nearby building. (a) Write an equation for the position of the shadow. (b) Write an equation for the velocity of the shadow.
To solve this problem it is necessary to apply the concepts related to the simple harmonic movement, to the speed in terms of displacement and the timpo, as well as the angular frequency and the period of frequency.
PART A) According to the description given, 5 revolutions are made in one minute (or 60 seconds) that is to say that the frequency would be given by
[tex]f = \frac{1}{12s^{-1}}[/tex]
Therefore the angular velocity can be found as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 0.52rad/s[/tex]
The displacement that determines the maximum displacement based on the angular velocity time and the time in the simple harmonic movement, is equal to the radius of the circle, in other words
[tex]x = Acos(\omega t)[/tex]
Where,
A = Amplitude
[tex]\omega =[/tex] Angular velocity
t = time
If for our given values the value of the amplitude is 2m and the value of the angular velocity is 0.52rad/s
[tex]x = 2cos(0.52t)[/tex]
So the equation for the position of the shadow is of
[tex]x(t) = 2cos(0.52t)[/tex]
PART B) The equation for the velocity of the shadow is calculated as a expression of the displacement against the time, if we differenciate the previous value found, we have that,
[tex]\frac{dx}{dt} = \frac{d(2cos(0.52t))}{dt}[/tex]
[tex]v(t) = 2(-0.52)sin(0.523t)[/tex]
[tex]v(t) = -1.05sin(0.52t)[/tex]
The position of the shadow can be described using the equation x=2cos(10πt+φ) and the velocity of the shadow can be described using the equation v=-2*10π*sin(10πt+φ), using principles of simple harmonic motion.
Explanation:The subject of this question involves physics and in particular, kinematics and rotational motion, as it relates to the shadow produced by a student standing on a rotating merry-go-round.
The position and velocity of the shadow will vary as the merry-go-round rotates, and these can be represented using mathematical equations. Let's denote the angular speed of the merry-go-round as 'w', which is given by the formula w = 2πn, where n is the number of rotations per minute. For the given question, n = 5 rotations per minute. Therefore, w = 10π rad/min.
(a) Position of the shadow: We can describe the position of the shadow in terms of simple harmonic motion - as the merry-go-round rotates, the shadow produced on the nearby building would be oscillating back and forth. This kind of motion can be mathematically represented as x=Acos(wt+φ), where x is the position of the shadow, A is the amplitude (which is the radius of the merry-go-round, hence 2 meters), w is the angular frequency, t is time, and φ is the phase constant.
(b) Velocity of the shadow: The velocity of the shadow is the derivative of the position with respect to time, which can be represented as v=-Aw*sin(wt+φ). In our case, the angular speed w = 10π rad/min, A is the amplitude (radius of the merry-go-round, hence 2 meters), and 't' is the time.
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A piece of charcoal used for cooking is found at the remains of an ancient campsite. A 1.09 kg sample of carbon from the wood has an activity of 2020 decays per minute. Find the age of the charcoal. Living material has an activity of 15 decays/minute per gram of carbon present and the half-life of 14C is 5730 y. Answer in units of y.
Answer:
t = 17199 years
Explanation:
given,
mass of sample = 1.09 Kg
Activity of living material = 15 decays / min /g
Activity of living material = 15 x 1000 decays /min /kg
Activity of living material per 1.09 kg A = 1.09 x 15 x 1000 decays / min
Activity of after time t is A ' = 2020
half life = 57300 years
desegregation constant
λ = 0.693 / 5700
[tex]A'= A e^{-\lambda\ t}[/tex]
[tex]A'= 1.09 \times 15 \times 1000 e^{-\lambda\ t}[/tex]
[tex]2020= 1.09 \times 15 \times 1000 e^{-\dfrac{0.693}{5700}\times t}[/tex]
[tex]0.124=e^{-\dfrac{0.693}{5700}\times t}[/tex]
taking ln both side
[tex]\dfrac{0.693}{5700}\times t = 2.09[/tex]
t = 17199 years
While jogging, a 70.0-kg student generates thermal energy at a rate of 1200 W. To maintain a constant body temperature of 37.0∘C, this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, irreversible body damage could occur. Protein structures in the body are irreversibly damaged if body temperature rises to 44.0∘C or above. The specific heat of a typical human body is 3480J/(kg⋅K), slightly less than that of water. (The difference is due to the presence of protein, fat, and minerals, which have lower specific heat capacities.)
Answer:
1421 seconds or 23.67 minutes
Explanation:
Q = Heat = 1200 W
c = Specific heat of human body = 3480 J/kgK
[tex]\Delta T[/tex] = Change in temperature = (44-37)°C
t = Time taken
[tex]Q=\dfrac{mc\Delta T}{t}\\\Rightarrow t=\dfrac{mc\Delta T}{Q}\\\Rightarrow t=\dfrac{70\times 3480\times (44-37)}{1200}\\\Rightarrow t=1421\ s[/tex]
The time student jogs before irreversible body damage occurs is 1421 seconds or 23.67 minutes
The body maintains a stable temperature through thermoregulation, with mechanisms like sweating playing an important role in removing heat. During physical activity, energy production increases and this heat needs to be dissipated to maintain body temperature. Failure of these mechanisms can lead to dangerous overheating.
Explanation:The phenomenon mentioned in the question deals with the concept of thermoregulation, which is the process through which the body maintains its core temperature within certain boundaries. When a person exercises, the body produces more heat as a byproduct of the energy utilised during physical activity. This heat is then removed by perspiration and other mechanisms to ensure body temperature remains within a safe range.
The energy generated (in this case, 1200 W) can be converted into heat and then removed from the body to maintain a constant body temperature. Sweat, as part of the body's thermoregulation system, plays a crucial role in this process. As sweat (which is water) evaporates from the skin, it takes a significant amount of heat energy with it, effectively cooling the body.
If these cooling mechanisms were ineffective and the body could not release this heat, it could cause a dangerous increase in body temperature, potentially leading to irreversible damage to proteins in the body if the temperature reaches 44°C or above. That's why adequate fluid intake is essential to replace the liquid lost through sweat and to prevent dehydration.
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Manufacturers of headache remedies routinely claim that their own brands are more potent pain relievers than the competing brands. Their way of making the comparison is to compare the number of molecules in the standard dosage. Tylenol uses 325 mg of acetaminophen (C8H9NO2) as the standard dose, while Advil uses 2.00 x 102 mg of ibuprofen (C13H18O2). Find the number of molecules of pain reliever in the standard doses of
(a) Tylenol and
(b) Advil.
a) 12.95*10^20 molecules
b) 5.84*10^20 molecules
Explanation:
a)Tylenol
Mass of Tylenol = 325 mg
(m)=325*10^-3 g
Molecular weight of Tylenol(M) = 151.16256 g/mol
the number of moles present
n = m/M
= 325*10^-3 g /151.1656 g/mol
= 2.15 *10^-3 mol
the number of molecules present
N= nNa
=(2.15*10^ -3 mol)(6.023*10^23 molecules/mol)
N=12.95*10 ^20 molecules
b)Advil
Mass of Advil = 200 mg
m=200*10^-3 g
Molecular weight of Advil(M) =206.28082 g/mol
the number of moles
n=m/M
=200*10^ -3 g/ 206.28082 g/mol
=0.96955*10^ -3 mol
the number of molecules resent
N=nNa
=(0.96955*10^ -3)(6.023*10^ 23 molecules/ mol)
N=5.84*10^20 molecules
Tylenol contains more molecules of active ingredient hence it is a more effective drug.
We have to find the number of molecules in each of the drugs. In the drug, tylenol, the molar mass of the compound C8H9NO2 is;
8(12) + 9(1) + 14 + 2(16) = 96 + 9 + 14 + 32 =151 g/mol
Number of moles = 325 × 10^-3 g/151 g/mol = 0.0022 moles
1 mole contains 6.02 × 10^23 molecules
0.0022 moles contains 0.0022 moles × 6.02 × 10^23 molecules /1 mole = 1.32 × 10^21 molecules
For Advil;
Molecular mass= 13(12) + 18(1) + 2(16) = 156 + 18 + 32 = 206 g/mol
Number of moles = 0.2 g/ 206 g/mol = 0.00097 moles
1 mole contains 6.02 × 10^23 molecules of advil
0.00097 moles contains 0.00097 moles × 6.02 × 10^23 molecules /1 mole = 5.8 × 10^20
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A 4-kg block is sliding down a plane as pictured, with the plane forming a 3-4-5 right triangle. Initially, the block is not moving. (However, we will assume only kinetic friction applies.) The coefficient of static friction between the block and the plane is 0.25. How long does it take for the block to reach the bottom of the plane? (Pick the answer closest to the true value.)A. 3.2 secondsB. 1.6 secondsC. 2.5 secondsD. 1.0 secondsE. 0.5 seconds
Answer:b
Explanation:
Given
mass of block [tex]m=4 kg[/tex]
coefficient of static friction [tex]\mu =0.25 [/tex]
height of triangle is [tex]h=3 m[/tex]
[tex]F_{net}=mg\sin \theta -\mu _kmg\cos \theta [/tex]
[tex]a_{net}=g\sin \theta -\mu _kg\cos \theta [/tex]
[tex]a_{net}=9.8\sin 37-0.25\times 9.8\times \cos 37[/tex]
[tex]a_{net}=5.897-1.956=3.94 m/s^2[/tex]
here [tex]s=5 m[/tex]
[tex]v^2-u^2=2 a_{net}s[/tex]
[tex]v=\sqrt{2\times 3.94\times 5}[/tex]
[tex]v=6.27 m/s[/tex]
time taken to reach bottom of plane
[tex]v=u+at[/tex]
[tex]6.27=0+3.94\times t[/tex]
[tex]t=1.59 s\approx 1.6 s[/tex]
On a bet, you try to remove water from a glass by blowing across the top of a vertical straw immersed in the water. What is the minimum speed you must give the air at the top of the straw to draw water upward through a height of 1.6cm?
Answer:
v₂ = 0.56 m / s
Explanation:
This exercise can be done using Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Where points 1 and 2 are on the surface of the glass and the top of the straw
The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0
The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m
We substitute in the equation
[tex]P_{atm}[/tex] + ρ g y₁ = [tex]P_{atm}[/tex] + ½ ρ v₂² + ρ g y₂
½ v₂² = g (y₂-y₁)
v₂ = √ 2 g (y₂-y₁)
Let's calculate
v₂ = √ (2 9.8 1.6 10⁻²)
v₂ = 0.56 m / s
A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm , y = 0, and then in another straight line from x = 40.0 mm , y = 0 to x = 40.0 mm , y = 55.0 mm . The wire is in an external uniform 0.300-T magnetic field in the +z direction, and the current through the wire is 5.10 A , directed from the origin into the wire.
A/Determine the magnitude of the magnetic force exerted by the external field on the wire.
B/Determine the direction of the magnetic force exerted by the external field on the wire measured clockwise from the positive x axis.
C/The wire is removed and replaced by one that runs directly from the origin to the locationx = 20.0mm , y = 35.0mm . If the current through this wire is also 4.90A , what is the magnitude of the magnetic force exerted by the external field on the wire?
D/The wire is removed and replaced by one that runs directly from the origin to the locationx = 20.0mm , y = 35.0mm . If the current through this wire is also 4.90A , what is the magnitude of the direction force exerted by the external field on the wire measured clockwise from the positive x axis.?
Answer:
0.1040512455 N
[tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
0.05925 N
[tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
Explanation:
I = Current
B = Magnetic field
Separation between end points is
[tex]l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm[/tex]
Effective force is given by
[tex]F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N[/tex]
The force is 0.1040512455 N
[tex]tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}[/tex]
The angle the force makes is given by
[tex]\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
The direction is [tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]
[tex]F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N[/tex]
The force is 0.05925 N
[tex]tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}[/tex]
[tex]\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
The direction is [tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]
A. Magnitude of the force: [tex]\( 0.104 \, \text{N} \)[/tex]
B. Direction of the force: [tex]\( 323.8^\circ \)[/tex]
C. Magnitude of the force: [tex]\( 0.0593 \, \text{N} \)[/tex]
D. Direction of the force: [tex]\( 330^\circ \)[/tex]
Part A: Determine the magnitude of the magnetic force exerted by the external field on the wire
Given: Length of wire segments: [tex]\( L_1 = 40.0 \, \text{mm} = 0.040 \, \text{m} \) and \( L_2 = 55.0 \, \text{mm} = 0.055 \, \text{m} \)[/tex]
- Magnetic field, [tex]\( B = 0.300 \, \text{T} \)[/tex]
- Current, [tex]\( I = 5.10 \, \text{A} \)[/tex]
The force on a current-carrying wire in a magnetic field is given by [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \).[/tex]
For the first segment along the x-axis:
- [tex]\( \mathbf{L}_1 = 0.040 \, \text{m} \, \hat{i} \)[/tex]
[tex]\( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]
[tex]\[ \mathbf{F}_1 = I (\mathbf{L}_1 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.040 \, \hat{i} \times 0.300 \, \hat{k}) \][/tex]
[tex]\[ \mathbf{F}_1 = 5.10 \times 0.040 \times 0.300 \, \hat{j} \][/tex]
[tex]\[ \mathbf{F}_1 = 0.0612 \, \text{N} \, \hat{j} \][/tex]
For the second segment along the y-axis:
[tex]- \( \mathbf{L}_2 = 0.055 \, \text{m} \, \hat{j} \)[/tex]
[tex]- \( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]
[tex]\[ \mathbf{F}_2 = I (\mathbf{L}_2 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.055 \, \hat{j} \times 0.300 \, \hat{k}) \][/tex]
[tex]\[ \mathbf{F}_2 = 5.10 \times 0.055 \times (-0.300) \, \hat{i} \][/tex]
[tex]\[ \mathbf{F}_2 = -0.08415 \, \text{N} \, \hat{i} \][/tex]
The net force [tex]\(\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2\)[/tex]:
[tex]\[ \mathbf{F} = -0.08415 \, \hat{i} + 0.0612 \, \hat{j} \][/tex]
The magnitude of the force is:
[tex]\[ |\mathbf{F}| = \sqrt{(-0.08415)^2 + (0.0612)^2} \][/tex]
[tex]\[ |\mathbf{F}| = \sqrt{0.00708 + 0.00375} \][/tex]
[tex]\[ |\mathbf{F}| = \sqrt{0.01083} \][/tex]
[tex]\[ |\mathbf{F}| \approx 0.104 \, \text{N} \][/tex]
Part B: The direction of the force can be found using the angle [tex]\( \theta \)[/tex] with respect to the positive x-axis:
[tex]\[ \theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) \][/tex]
[tex]\[ \theta = \tan^{-1} \left( \frac{0.0612}{-0.08415} \right) \][/tex]
[tex]\[ \theta = \tan^{-1} \left( -0.727 \right) \][/tex]
[tex]\[ \theta \approx -36.2^\circ \][/tex]
Since this angle is measured counterclockwise from the positive x-axis, the direction clockwise is:
[tex]\[ 360^\circ - 36.2^\circ = 323.8^\circ \][/tex]
Part C: Determine the magnitude of the magnetic force exerted by the external field on the wire
For the wire directly from the origin to [tex]\( x = 20.0 \, \text{mm}, y = 35.0 \, \text{mm} \):[/tex]
- Length [tex]\( L = \sqrt{(20.0 \, \text{mm})^2 + (35.0 \, \text{mm})^2} = \sqrt{(0.020 \, \text{m})^2 + (0.035 \, \text{m})^2} \)[/tex]
- Length [tex]\( L = \sqrt{0.0004 + 0.001225} = \sqrt{0.001625} \)[/tex]- Length [tex]\( L \approx 0.0403 \, \text{m} \)[/tex]
[tex]\[ \mathbf{L} = 0.0403 \, \text{m} \][/tex]
Given current [tex]\( I = 4.90 \, \text{A} \)[/tex]:
[tex]\[ \mathbf{F} = I L B \sin(\theta) \][/tex]
[tex]\[ \theta = \angle \text{between } \mathbf{L} \text{ and } \mathbf{B} = 90^\circ \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times 0.0403 \times 0.300 \][/tex]
[tex]\[ \mathbf{F} \approx 0.0593 \, \text{N} \][/tex]
Part D: - Current: [tex]\( 4.90 \, \text{A} \)[/tex]
- Magnetic field: [tex]\( 0.300 \, \text{T} \, \hat{k} \)[/tex]
- Displacement vector: [tex]\( \mathbf{L} = 0.020 \, \hat{i} + 0.035 \, \hat{j} \)[/tex]
- Force vector: [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \)[/tex]
[tex]\[ \mathbf{L} \times \mathbf{B} = (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]
[tex]\[ \mathbf{F} = 4.90 (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times (0.020 \times 0.300 \, \hat{j} - 0.035 \times 0.300 \, \hat{i}) \][/tex]
[tex]\[ \mathbf{F} = 4.90 \times (0.006 \, \hat{j} - 0.0105 \, \hat{i}) \][/tex]
[tex]\[ \mathbf{F} = 0.0294 \, \hat{j} - 0.05145 \, \hat{i} \][/tex]
The magnitude of the force:
[tex]\[ |\mathbf{F}| = \sqrt{(0.0294)^2 + (-0.05145)^2} \approx 0.0593 \, \text{N} \][/tex]
The angle with respect to the x-axis:
[tex]\[ \theta = \tan^{-1} \left( \frac{0.0294}{-0.05145} \right) = \tan^{-1} \left( -0.571 \right) \approx -30^\circ \][/tex]
The direction clockwise is:
[tex]\[ 360^\circ - 30^\circ = 330^\circ \][/tex]
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is empty, aside from containing air. The bucket is then inverted so that its open end is down and, being careful not to lose any of the air trapped inside, the bucket is lowered below the surface of a fresh-water lake so the water-air interface in the bucket is 20.0 m below the surface of the lake.
To keep the calculations simple, use g = 10 m/s2, atmospheric pressure = 1. 105 Pa, and the density of water as 1000 kg/m3.a. If the temperature is the same at the surface of the lake and at a depth of 20.0 m below the surface, what is the height of the cylinder of air in the bucket when the bucket is at a depth of 20.0 m below the surface of the lake?b. If, instead, the temperature changes from 300 K at the surface of the lake to 275 K at a depth of 20.0 m below the surface of the lake, what is the height of the cylinder of air in the bucket?
Answer:
12 cm
Explanation:
[tex]P_1[/tex] = Initial pressure = [tex]P_a=1\times 10^5\ Pa[/tex]
[tex]P_2[/tex] = Final pressure = [tex]P_a+\rho_w gh[/tex]
h = Depth of cylinder = 36 cm
g = Acceleration due to gravity = 10 m/s²
[tex]\rho_w[/tex] = Density of water = 1000 kg/m³
[tex]h_1[/tex] = Depth of lake = 20 m
From the ideal gas relation we have
[tex]P_1V_1=P_2V_2\\\Rightarrow P_a(\pi r^2h)=(P_a+\rho_w gh_1)\pi r^2h'\\\Rightarrow 1\times 10^5\times 36=(1\times 10^5+1000\times 10\times 20)h'\\\Rightarrow h'=\dfrac{1\times 10^5\times 36}{1\times 10^5+1000\times 10\times 20}\\\Rightarrow h'=12\ cm[/tex]
The height of the cylinder of air in the bucket when the bucket is at the given depth is 12 cm
A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.Part AFind the work done by the gas during the initial expansion.W initial = JPart BFind the heat added to the gas during the initial expansion.Q initial = JPart CFind internal-energy change of the gas during the initial expansion.ΔU initial = JPart DFind the work done during the final cooling;W final = J
Answer:
A. [tex]W=600\ J[/tex]
B. [tex]Q=2112\ J[/tex]
C. [tex]\Delta U=1512\ J[/tex]
D. [tex]W=0\ J[/tex]
Explanation:
Given:
no. of moles of oxygen in the cylinder, [tex]n=0.2[/tex]initial pressure in the cylinder, [tex]P_i=2\times 10^5\ Pa[/tex]initial temperature of the gas in the cylinder, [tex]T_i=360\ K[/tex]According to the question the final volume becomes twice of the initial volume.
Using ideal gas law:
[tex]P.V=n.R.T[/tex]
[tex]2\times 10^5\times V_i=0.2\times 8.314\times 360[/tex]
[tex]V_i=0.003\ m^3[/tex]
A.
Work done by the gas during the initial isobaric expansion:
[tex]W=P.dV[/tex]
[tex]W=P_i\times (V_f-V_i)[/tex]
[tex]W=2\times 10^5\times (0.006-0.003)[/tex]
[tex]W=600\ J[/tex]
C.
we have the specific heat capacity of oxygen at constant pressure as:
[tex]c_v=21\ J.mol^{-1}.K^{-1}[/tex]
Now we apply Charles Law:
[tex]\frac{V_i}{T_i} =\frac{V_f}{T_f}[/tex]
[tex]\frac{0.003}{360} =\frac{0.006}{T_f}[/tex]
[tex]T_f=720\ K[/tex]
Now change in internal energy:
[tex]\Delta U=n.c_p.(T_f-T_i)[/tex]
[tex]\Delta U=0.2\times 21\times (720-360)[/tex]
[tex]\Delta U=1512\ J[/tex]
B.
Now heat added to the system:
[tex]Q=W+\Delta U[/tex]
[tex]Q=600+1512[/tex]
[tex]Q=2112\ J[/tex]
D.
Since during final cooling the process is isochoric (i.e. the volume does not changes). So,
[tex]W=0\ J[/tex]
Water has a specific heat capacity nearly nine times that of iron. Suppose a 50-g pellet of iron at a temperature of 200∘C is dropped into 50 g of water at a temperature of 20∘C.
When the system reaches thermal equilibrium, its temperature will be
a. closer to 20∘C.
b. halfway between the two initial temperatures.
c. closer to 200∘C.
Answer:
a. closer to 20∘C
Explanation:
[tex]m_{p}[/tex] = mass of pallet = 50 g = 0.050 kg
[tex]c_{p}[/tex] = specific heat of pallet = specific heat of iron
[tex]T_{pi}[/tex] = Initial temperature of pellet = 200 C
[tex]m_{w}[/tex] = mass of water = 50 g = 0.050 kg
[tex]c_{w}[/tex] = specific heat of water
[tex]T_{wi}[/tex] = Initial temperature of water = 20 C
[tex]T_{e}[/tex] = Final equilibrium temperature
Also given that
[tex]c_{w} = 9 c_{p} [/tex]
Using conservation of energy
Energy gained by water = Energy lost by pellet
[tex]m_{w} c_{w} (T_{e} - T_{wi}) = m_{p} c_{p} (T_{pi} - T_{e})\\(0.050) (9) c_{p} (T_{e} - 20) = (0.050) c_{p} (200 - T_{e})\\ (9) (T_{e} - 20) = (200 - T_{e})\\T_{e} = 38 C[/tex]
hence the correct choice is
a. closer to 20∘C
Final answer:
When a 50-g iron pellet at 200°C is dropped into 50 g of water at 20°C, the final temperature of the system upon reaching thermal equilibrium will be closer to 20°C. This is due to water's higher specific heat capacity, causing it to undergo a smaller temperature change during heat transfer.
Explanation:
The question involves a calorimetry problem, where two objects at different temperatures - a 50-g pellet of iron at 200°C and 50 g of water at 20°C - are combined and allowed to reach thermal equilibrium. Given that water has a specific heat capacity that is significantly higher than that of iron, the thermal energy will transfer from the iron to the water until both reach the same temperature.
As the water has a higher specific heat capacity, it will undergo a smaller change in temperature for a given amount of heat transfer. Hence, when the iron and water achieve thermal equilibrium, the final temperature will be closer to 20°C than to 200°C. This is because the 50 g of water will absorb more heat from the iron pellet with a less significant rise in temperature, compared to the temperature drop that iron experiences.
Based on the principles of heat transfer and specific heat capacity, the correct answer to the student's question is: a. closer to 20°C.
A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 = 4ft/s. Knowing that the tape does not slip on the drums, determine (a) the angular acceleration of drum B, and (b) the number of revolutions executed by drum B during the 4-s interval. 5) A rocket weighs 2600 lb, including 2200 lb of fuel, which is consumed at the rate of 25 lb/s and ejected with a relative velocity of 13000 ft/s. Knowing that the rocket is fired vertically from the ground, determine its acceleration (a) as it is fired, and (b) as the last particle of fuel is being consumed.
Answer:
Question 1)
a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.
[tex]v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2[/tex]
This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,
[tex]v = \omega R\\a = \alpha R[/tex]
where α is the angular acceleration.
In order to continue this question, the radius of the drums should be given.
Let us denote the radius of the drums as R, the angular acceleration of drum B is
α = 0.5/R.
b) The distance travelled by the drums can be found by the following kinematics formula:
[tex]v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft[/tex]
One revolution is equal to the circumference of the drum. So, total number of revolutions is
[tex]x / (2\pi R) = 6/(\pi R)[/tex]
Question 2)
a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:
[tex]a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2[/tex]
b) [tex]a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2[/tex]
The angular acceleration of drum B is 0.5 ft/s² and the number of revolutions executed by drum B during the 4-s interval is 6 ft.
Explanation:To determine the angular acceleration of drum B, we can use the formula:
α = (v1 - v0) / t
Where α is the angular acceleration, v1 and v0 are the final and initial speeds respectively, and t is the time interval.
Substituting the given values, we have:
α = (4 ft/s - 2 ft/s) / 4 s = 0.5 ft/s²
To find the number of revolutions executed by drum B during the 4 s interval, we can use the formula:
θ = (ω0 + ω1) / 2 * t
Where θ is the angle in radians, ω0 and ω1 are the initial and final angular velocities respectively, and t is the time interval.
Since the tape does not slip on the drums, the angular velocity of drum B is the same as the linear velocity of the tape. Thus, ω0 = v0 and ω1 = v1.
Substituting the given values, we have:
θ = (2 ft/s + 4 ft/s) / 2 * 4 s = 6 ft
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If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 55 MN/m , determine the displacement of end A when the 60-kN force is applied. Take E = 200 GPa.
The question pertains to calculating the displacement at end A of a steel rod with a spring attached when subject to a specific force, using concepts of mechanical engineering such as Hooke's law and elastic deformation.
Explanation:The student's question involves calculating the displacement of end A when a 60-kN force is applied to a 20-mm-diameter rod made of A-36 steel which has a spring attached with a stiffness k = 55 MN/m and taking into account that the modulus of elasticity E = 200 GPa. This problem is a straightforward application of Hooke's law combined with the elastic deformation formula (stress = force/area, strain = change in length/original length, and Hooke's law: F = kx for springs), which are part of mechanical engineering and physics topics on material strength and deformation.
To find the displacement, we should consider the deformation of the rod under the applied force and the compression of the spring separately. The rod's deformation can be found using the modulus of elasticity E and the cross-sectional area derived from the diameter, while the spring's compression is directly related to the force applied and the spring constant k.
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A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put on the launcher, the launcher spring compresses by 10.0 cm. The watermelon is then pushed down by an additional 30.0 cm and it’s ready to go. Just before the launch, how much energy is stored in the spring?
To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.
Since the forces are balanced the Spring force is equal to the force of the weight that is
[tex]F_s = F_g[/tex]
[tex]kx = mg[/tex]
Where,
k = Spring constant
x = Displacement
m = Mass
g = Gravitational Acceleration
Re-arrange to find the spring constant
[tex]k = \frac{mg}{x}[/tex]
[tex]k = \frac{8*9.8}{0.1}[/tex]
[tex]k = 784N/m[/tex]
Just before launch the compression is 40cm, then from Potential Elastic Energy definition
[tex]PE = \frac{1}{2} kx^2[/tex]
[tex]PE =\frac{1}{2} 784*0.4^2[/tex]
[tex]PE = 63.72J[/tex]
Therefore the energy stored in the spring is 63.72J
Long Snorkel: Inhalation of a breath occurs when the muscles surrounding the human lungs move to increase the volume of the lungs, thereby reducing the air pressure in the lungs. The difference between the reduced pressure in the lungs and outside atmospheric pressure causes a flow of air into the lungs. The maximum reduction in air pressure that muscles of the chest can produce in the lungs against the surrounding air pressure on the chest and body is about 3740 Pa. Consider the longest snorkel that a human can operate. The minimum pressure difference needed to take in a breath is about 188 Pa. Find the depth h that a person could swim to and still breathe with this snorkel
Answer:
h = 0.362 m
Explanation:
The pressure equation with depth is
P₂ = [tex]P_{atm}[/tex] +ρ g h
The gauge pressure is
P2 - [tex]P_{atm}[/tex] = ρ g h
This is the pressure that muscles can create
P₂ - [tex]P_{atm}[/tex]= 3740 Pa
But still the person needs a small pressure for the transfer of gases, so
P₂ - [tex]P_{atm}[/tex] = 3740 - 188 = 3552 Pa
This is the maximum pressure difference, where the person can still breathe,
Let's clear the height
h = 3552 / ρ g
h = 3552 / (1000 9.8)
h = 0.362 m
This is the maximum depth where the person can still breathe normally.
Final answer:
The maximum depth that a person could swim to and still breathe with the snorkel is approximately 1.9 cm.
Explanation:
To find the depth that a person could swim to and still breathe with the snorkel, we need to consider the pressure difference needed for inhalation. The maximum reduction in air pressure that the chest muscles can produce in the lungs is 3740 Pa. The minimum pressure difference needed for inhalation is 188 Pa. Therefore, the maximum depth h that a person could swim to and still breathe with the snorkel is determined by the difference in atmospheric pressure at the surface and the pressure at that depth.
Using the equation for gauge pressure, we can calculate the maximum depth:
Pgauge = ρgh
Where:
Pgauge is the gauge pressure
ρ is the density of the fluid (water)
g is the acceleration due to gravity
h is the depth
Given that the minimum pressure difference is 188 Pa, we can rearrange the equation to solve for h:
h = Pgauge / (ρg)
Using the values for the density of water (1000 kg/m³) and acceleration due to gravity (9.8 m/s²), we can calculate the maximum depth:
h = 188 Pa / (1000 kg/m³ * 9.8 m/s²) = 0.019 m = 1.9 cm
Therefore, a person could swim to a maximum depth of approximately 1.9 cm and still breathe with the snorkel.
A statue of unknown volume and density is suspended from a string. When suspended in air, the tension in the string is Tair; however, when submerged in water, the tension is Tw
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way in which this is done is to seat them in a chair of known mass attached to a spring of known force constant and measure the period of the oscillations of this system. The 36.2kg chair alone oscillates with a period of 1.15s , and the period with the astronaut sitting in the chair is 2.20s
Part A: Find the force constant of the spring.
Part B: Find the mass of the astronaut.
Answer:
a. K = 1080.61 N/m
b. mₐ = 96.28 kg
Explanation:
T = 2π * √ m / K
T₁ = 2π * √m₁ / K , T₂ = 2π * √m₂ / K
m₁ = 36.2 kg , m₂ = m₁ + mₐ
T₁ = 1.15 s , T₂ = 2.2 s
Equal period to determine the force constant of the spring
a.
K = 4π²* m₁ / T₁²
K = 4π² * 36.2 kg / 1.15² s
K = 1080.61 N/m
So replacing and solve can find m₂
b.
m₂ = T₂² * m₁ / T₁² ⇒ m₂ = 2.20² s * 36.2 kg / 1.15 ² s
m₂ = 132.48
mₐ = m₂ - m₁
mₐ = ( 132.48 - 36.2 ) kg = 96.28 kg
Twenty grams of a solid at 70°C is place in 100 grams of a fluid at 20°C. Thermal equilibrium is reached at 30°C.
The specific heat of the solid:
a. is equal to that of the fluid.
b. is less than that of the fluid.
c. is more than that of the fluid.
d. cannot be compared to that of a material in a different phase.
Answer:
c. is more than that of the fluid.
Explanation:
This problem is based on the conservation of energy and the concept of thermal equilibrium
[tex]heat= m s \Delta T [/tex]
m= mass
s= specific heat
\DeltaT=change in temperature
let s1= specific heat of solid and s2= specific heat of liquid
then
Heat lost by solid= [tex]20(s_1)(70-30)=800s_1 [/tex]
Heat gained by fluid=[tex]100(s_2)(30-20)=1000s_2 [/tex]
Now heat gained = heat lost
therefore,
1000 S_2=800 S_1
S_1=1.25 S_2
so the specific heat of solid is more than that of the fluid.
The correct option is:
(c) is more than that of the fluid.
Conservation of energy:
The conservation of energy suggests that the heat energy lost by the solid must be equal to the heat energy gained by the liquid.
Let the specific heat of the solid be [tex]s_s[/tex] and the specific heat of the liquid be [tex]s_l[/tex].
Heat energy lost by the solid is given by:
[tex]\Delta Q_s=ms_s\Delta T\\\\ \Delta Q_s=20s_s(70-30)\\\\ \Delta Q_s=800s_s[/tex]
Heat energy gained by the liquid:
[tex]\Delta Q_l=ms_l\Delta T\\\\ \Delta Q_l=100s_l(30-20)\\\\ \Delta Q_l=1000s_l[/tex]
According to the conservation of energy:
[tex]\Delta Q_s=\Delta Q_l\\\\800s_s=1000s_l\\\\\frac{s_s}{s_l}=\frac{1000}{800}\\\\\frac{s_s}{s_l}=\frac{5}{4}[/tex]
Hence, the specific heat of the solid is more than that of the fluid
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Wire resistor A has twice the length and twice the cross sectional area of wire resistor B. Which of the following accurately compares the resistances of wire resistors A and B?a) Wire A has twice the resistance of wire B.b) Wire A has half the resistance of wire B.c) Wire A has the same resistance as wire B.d) None of the above
Answer: Option (c) is the correct answer.
Explanation:
It is known that the relation between resistance, length and cross-sectional area is as follows.
R = [tex]\rho \frac{l}{A}[/tex]
Let the resistance of resistor A is denoted by R and the resistance of resistor B is denoted by R'.
Hence, for resistor A the expression for resistance according to the given data is as follows.
R = [tex]\rho \frac{2l}{2A}[/tex]
On cancelling the common terms we get the expression as follows.
R = [tex]\rho \frac{l}{A}[/tex]
Now, the resistance for resistor B is as follows.
R' = [tex]\rho \frac{l'}{A'}[/tex]
Thus, we can conclude that the statement, Wire A has the same resistance as wire B, accurately compares the resistances of wire resistors A and B.
A light ray enters a glass enclosed fish tank. From air it enters the glass at 20.° with respect to the surface, then emerges into the water. The index for glass is 1.50 and for water 1.33.
(a) What is the angle of refraction in the glass?
(b) What is the angle of refraction in the water?
(c) Is there any incident angle in air for which the ray will not enter the water due to total internal reflection?
Answer:
a) [tex]\angle r_{ag}=38.79^{\circ}[/tex]
b) [tex]\angle r_{gw}=44.95^{\circ}[/tex]
c) not possible
Explanation:
Given:
angle of incidence on the air-glass interface, [tex]\angle i_{ag}=90-20=70^{\circ}[/tex]
refractive index of glass with respect to air, [tex]n_g=1.5[/tex]
refractive index of water with respect to air, [tex]n_a=1.33[/tex]
a)
For angle of refraction in glass we use Snell's law:
[tex]n_g=\frac{sin\ i_{ag}}{sin\ r{ag}}[/tex]
[tex]1.5=\frac{sin\ 70}{sin\ r_{ag}}[/tex]
[tex]\angle r_{ag}=38.79^{\circ}[/tex]
b)
Now we have angle of incident for glass-water interface, [tex]\angle i_{gw}=\angle r_{ag}=38.79^{\circ}[/tex]
And the refractive index of water with respect to glass:
[tex]n_{gw}=\frac{n_w}{n_g}[/tex]
[tex]n_{gw}=\frac{1.33}{1.5}[/tex]
[tex]n_{gw}=0.8867 [/tex]
Therefore, angle of refraction in the water:
[tex]n_{gw}=\frac{sin\ i_{gw}}{sin\ r_{gw}}[/tex]
[tex]0.8867 =\frac{sin\ 38.79}{sin\ r_{gw}}[/tex]
[tex]\angle r_{gw}=44.95^{\circ}[/tex]
c)
For total internal reflection through water the light must enter the glass-water interface at an angle greater than the critical angle.
So,
[tex]n_{gw}=\frac{sin\ i_{(gw)_c}}{sin\ 90}[/tex]
[tex]0.8867 =\frac{sin\ i_{(gw)_c}}{1}[/tex]
[tex]i_{(gw)_c}=62.46^{\circ}[/tex]
Now this angle will become angle of refraction for the air-glass interface.
Hence,
[tex]n_g=\frac{sin\ i_{ag}}{sin\ i_{(gw)_c}}[/tex]
[tex]1.5=\frac{sin\ i_{ag}}{0.8867 }[/tex]
[tex]sin\ i_{ag}=1.33005[/tex]
Since we do not have any angle for sine for which the value exceeds 1, therefore it is not possible that the light will reflect from the water.
A Michelson interferometer operating at a 600nm wavelength has a 2.02-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028. How many bright-dark-bright fringe shifts are observed as the cell fills with air? Am My Answers Give Up Submit
Answer:
19
Explanation:
d = Length of cell = 2.02 cm
[tex]\lambda[/tex] = Wavelength = 600 nm
n = Refractive index of air = 1.00028
In the Michelson interferometer the relation of the number of the bright-dark-bright fringe is given by
[tex]\Delta m=\dfrac{2d}{\lambda}(n-1)\\\Rightarrow \Delta m=\dfrac{2\times 2.02\times 10^{-2}}{600\times 10^{-9}}\times (1.00028-1)\\\Rightarrow \Delta m=18.853\approx 19[/tex]
The number of bright-dark-bright fringe shifts that are observed as the cell fills with air are 19
Which of the following experiments could be used to determine the inertial mass of a block? A. Place the block on a rough horizontal surface. Lift one end of the surface up and measure the angle the surface makes with the horizontal at the moment the block begins to slide. B. A Drop the block from different heights and measure the time of fall from each height. C. Place the block on a rough horizontal surface. Give the block an initial velocity and then let it come to rest. Measure the initial velocity and the distance the block moves in coming to rest. D. Use a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force.
Answer:
D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force
Explanation:
For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.
The experiment that could be used to determine the inertial mass of a block is ; ( D ) Using a spring scale to exert a force on the block.Measure the acceleration of the block and the applied force.
Inertial mass is the mass of a body that poses an inertial resistance to the acceleration of a body when forces are applied to it. before you inertial mass can be determined, force must be applied to the body at rest .
The experiment that is suitable to determine the inertial mass is using a spring scale to apply a force on the block, then take measurement of the acceleration experienced by the block and amount of force applied.
Inertial mass = acceleration / net force applied
Hence we can conclude that The experiment that could be used to determine the inertial mass of a block is Using a spring scale to exert a force on the block.Measure the acceleration of the block and the applied force.
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Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.
If the temperature difference is 20° C from the inside of the house to the outside air, what is the rate of heat flow through this window?
(Thermal conductivity for glass is 0.84 J/s⋅m⋅°C and for air 0.023 4 J/s⋅m⋅°C.)
a. 7 700 Wb. 1 900 Wc. 547 Wd. 180 W
To solve this problem it is necessary to apply the concepts related to rate of thermal conduction
[tex]\frac{Q}{t} = \frac{kA\Delta T}{d}[/tex]
The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, [tex]\Delta T[/tex], T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.
The change made between glass and air would be determined by:
[tex](\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}[/tex]
[tex] k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}[/tex]
[tex]\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}[/tex]
[tex]\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}[/tex]
[tex]\Delta T_{air} = 44.9 \Delta T_{glass}[/tex]
There are two layers of Glass and one layer of Air so the total temperature would be given as,
[tex]\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}[/tex]
[tex]\Delta T = 2\Delta T_{glass} +\Delta T_{air}[/tex]
[tex]20\°C = 46.9\Delta T_{glass}[/tex]
[tex]\Delta T_{glass} = 0.426\°C[/tex]
Finally the rate of heat flow through this windows is given as,
[tex]\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}[/tex]
[tex]\Delta {Q}{t} = 0.84*24*10 -3*0.426[/tex]
[tex]\Delta {Q}{t} = 179W[/tex]
Therefore the correct answer is D. 180W.
A uniform ladder of mass (m) and length (L) leans against a frictionless wall. If the coefficient of static friction between the ladder and the ground is 0.36, what is the minimum angle (θ) between the ladder and the floor at which the ladder will not slip?
Newton's second law for linear and rotational motion allows us to find the minimum angle so that the ladder does not slide is:
θ= 54.2º
Newton's second law for stable rotational motion that torque is equal to the product of the moment of inertia times the angular acceleration, when the angular acceleration is zero we have an equilibrium condition.
Σ τ = 0
τ = F r sin θ
Where θ is the torque, F the force, (r sin θ) the perpendicular distance.
In the attached we have a free-body diagram of the forces.
x- axis
R - fr = 0
y -axis
N - W = 0
N = W
The friction force has the expression.
fr = μ N
we substitute
R = μ W (1)
We use Newton's second rotational law where the pivot point is the base of the ladder.
R y - W x = 0
Let's use trigonometry.
sin θ = y / L
cos θ = x / L
y = L sin θ
x = L cos θ
Let's substitute.
R L sin θ = [tex]W \frac{L}{2} cos \theta[/tex]
R sin θ = [tex]\frac{1}{2} W cos \theta[/tex]
We use equation 1.
μ W sin θ =[tex]\frac{1}{2}[/tex] W cos θ
tan θ = [tex]\frac{1}{2 \mu }[/tex]
θ = tan⁻¹ [tex]\frac{1}{2 \mu}[/tex]
We calculate.
θ = tan⁻¹ [tex]\frac{1 }{2 \ 0.36}[/tex]
θ = 54.2º
In conclusion, using Newton's second law we can find the minimum angle for the ladder not to slide is:
tae 54.2º
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The minimum angle between the ladder and the floor at which the ladder will not slip is approximately 21.8 degrees.
Explanation:To determine the minimum angle (θ) between the ladder and the floor at which the ladder will not slip, we need to consider the forces acting on the ladder. These forces include the normal reaction force (N) from the floor, the static friction force (f) between the ladder and the ground, and the weight of the ladder (w). The ladder will not slip as long as the net torque acting on it is zero, which occurs when the static friction force is greater than or equal to the force component that wants to make the ladder slip.
In this case, the force component that wants to make the ladder slip is the horizontal component of the ladder's weight, which is w*sin(θ). The static friction force can be calculated using the equation f = μsN, where μs is the coefficient of static friction and N is the normal reaction force.
We can set up an inequality to find the minimum angle: μsN ≥ w*sin(θ). Substituting the values given in the problem, we have 0.36*N ≥ w*sin(θ). Dividing both sides by N gives us 0.36 ≥ sin(θ). Taking the inverse sine of both sides gives us θ ≥ sin-1(0.36).
Therefore, the minimum angle between the ladder and the floor at which the ladder will not slip is approximately 21.8 degrees.
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A container of water, diameter 12 cm, has a small opening near the bottom that can be unplugged so that the water can run out. If the top of the tank is open to the atmosphere, what is the exit speed of the water leaving through the hole. The water level is 15 cm above the bottom of the container. The center of the 3.0 diameter hole is 4.0 cm from the bottom. How long does it take the water to hit the ground? How far from the container will the water land?
Answer:
1.32 m
Explanation:
from the given figure we can find the velocity at the hole in the container
let it be v and we know that
then [tex]v= \sqrt{2gh}[/tex]
h= 15-4= 11 cm , g=9.81
[tex]v= \sqrt{2(9.81)(11)}[/tex]
v=14.69 m/s
now using
s=ut+0.5at^2
t= is the time required by water to reach bottom of the container
u = velocity in vertical direction is zero.
therefore,
[tex]0.04= 0\times t+\frac{1}{2}\times9.81\times t^2[/tex]
t= 0.09030 sec
let x be the distance far from the container will the water land
x=vt
x=14.69×0.09030 = 1.32 m
Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and place it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6370 km.) Ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation.
To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.
Mathematically the conservation of these two energies can be given through
[tex]W = U_f - U_i[/tex]
Where,
W = Work
[tex]U_f =[/tex] Final gravitational Potential energy
[tex]U_i =[/tex] Initial gravitational Potential energy
When the spacecraft of mass m is on the surface of the earth then the energy possessed by it
[tex]U_i = \frac{-GMm}{R}[/tex]
Where
M = mass of earth
m = Mass of spacecraft
R = Radius of earth
Let the spacecraft is now in an orbit whose attitude is [tex]R_{orbit} \approx R[/tex] then the energy possessed by the spacecraft is
[tex]U_f = \frac{-GMm}{2R}[/tex]
Work needed to put it in orbit is the difference between the above two
[tex]W = U_f - U_i[/tex]
[tex]W = -GMm (\frac{1}{2R}-\frac{1}{R})[/tex]
Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is
[tex]W = \frac{GMm}{2R}[/tex]
How much heat energy is required to vaporize a 1.0-g ice cube at 0°C? The heat of fusion of ice is 80 cal/g.
The heat of vaporization of water is 540 cal/g, and cwater = 1.00 cal/g⋅°C.
a. 620 cal
b. 720 cal
c. 820 cal
d. 1 kcal
The heat required to convert 1.0-g of ice cube at 0°C to vapor is 720 cal.
We know that when heat is supplied to a substance, change of state occurs as the object moves from a particular state of matter to another. The heat required to convert 1.0-g of ice cube at 0°C to vapor is obtained from;
H = mLfus + mcdT + mLvap
H = (1 g * 80 cal/g) + (1 g * 1.00 cal/g⋅°C * (100 - 0)) + (1 g * 540 cal/g)
H = 80 cal + 100 cal + 540 cal
H = 720 cal
The heat required to convert 1.0-g of ice cube at 0°C to vapor is 720 cal.
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The total amount of heat required to vaporize a 1-g ice cube at 0°C considering the stages of heat of fusion, heating up water, and heat of vaporization is 720 cal.
Explanation:To calculate the total amount of heat energy required to vaporize a 1.0-g ice cube at 0°C, we need to consider the heat added to turn the ice into water (heat of fusion), heat to raise the temperature of that water from 0°C to 100°C, and finally heat to turn the water into steam (heat of vaporization).
First, we use the heat of fusion of ice, which is 80 cal/g, thus the heat needed to turn 1g of ice into water at 0°C is 1g * 80 cal/g = 80 cal.
Secondly, to raise the water temperature from 0 to 100°C, we use cwater = 1.00 cal/g⋅°C, thus the required heat is 1g * 100°C * 1.00 cal/g⋅°C = 100 cal.
Finally, we use the heat of vaporization of water which is 540 cal/g, for 1g of water, the heat needed is 1g * 540 cal/g = 540 cal.
Adding all those up give us a total heat of 80 cal + 100 cal + 540 cal = 720 cal.
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A rigid rod is rotating at a constant angular speed about an axis that is perpendicular to one end of the rod. A small ball is fixed to the rod at a distance that is 12 cm from the axis. The centripetal force acting on this ball is 1.7 N. What would be the centripetal force that acts on the ball if the ball were fixed to the rod at a distance of 33 cm from the axis?
Answer:
Fc₂ = 4.675 N
Explanation:
r₁ = 12 cm = 0.12 m
Fc= 1.7 N
Fc = mV²/r = m w² r ( ∴ V= r w )
Fc = m w² r ----------------------- (1)
As angular speed (w) and mass are constant.
let m * w² = k ------------------(2)
Put equation (2) in equation (1).
⇒ Fc = m w² r = k * r ( ∴ m w² = k)
⇒ Fc =k * r ----------------------( 3)
⇒ Fc₁ =k * r₁ = 1.7 N
⇒ k = 1.7/ r = 1.7 / 0.12 N ( as r₁=0.12 m )
⇒ k = 14.17
Centripetal force hen the rod is fixed at 33 cm from the axis:
From equation ( 3 )
Fc₂ = k * r₂
Fc₂ = 14.17 * 0.33 N ( ∴ r₂ = 0.33 m)
Fc₂ = 4.675 N