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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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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You drop an egg off a bridge. What forces act on the egg as it falls?
!! Brainliest to correct answer(s) !!
Answer:
Some of the Forces that are related to to the gg dropping are Grvaity, air resistance/ Air friction! hope this helps . Reley on Newtons third Law!! Also use The words Compresion and Velocity in your report!! :) LMK IF THIS HELPED>
Explanation:
When an egg is dropped off a bridge, the main forces acting on it are gravity, air resistance, and, if it falls into water, the buoyant force.
Explanation:When an egg is dropped off a bridge, several forces act on it as it falls. The main force acting on the egg is gravity, which pulls the egg downwards towards the ground. Another force is air resistance, which opposes the motion of the falling egg and slows it down. Additionally, there may be a buoyant force acting on the egg if it falls into water, which pushes the egg upwards. Gravity is the force that gives weight to objects and pulls them towards the center of the Earth. It is responsible for the downward motion of the egg.
Air resistance is the frictional force exerted by the air on the falling egg. It increases with the speed and surface area of the falling object. Buoyant force, on the other hand, is an upward force exerted by a fluid, such as water, on an object partially or fully submerged in it. If the egg falls into water, the buoyant force would act on it, partially counteracting the force of gravity. Overall, the forces acting on the egg as it falls off a bridge are gravity, air resistance, and the buoyant force (if it falls into water).
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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]
An automobile (and its occupants) of total mass M = 2000 kg, is moving through a
curved dip in the road of radius R = 20 m at a constant speed v = 20 m/s. For this analysis, you can
neglect air resistance. Consider the automobile (and its occupants) as the system of interest. Use g =
10 m/s2.
Calculate the normal force exerted by the road on the system (car and its occupants).
A) 60,000 N
B) 20,000 N
C) 40,000 N
D) 50,000 N
E) 30,000 N
Answer:
Normal force, N = 60000 N
Explanation:
It is given that,
Mass of the automobile, m = 2000 kg
Radius of the curved road, r = 20 m
Speed of the automobile, v = 20 m/s
Let N is the normal and F is the net force acting on the automobile or the centripetal force. It is given by :
[tex]N-mg=\dfrac{mv^2}{r}[/tex]
[tex]N=\dfrac{mv^2}{r}+mg[/tex]
[tex]N=m(\dfrac{v^2}{r}+g)[/tex]
[tex]N=2000\times (\dfrac{(20)^2}{20}+10)[/tex]
N = 60000 N
So, the normal force exerted by the road on the system is 60000 Newton. Hence, this is the required solution.
Final answer:
The normal force exerted by the road on the system (car and its occupants) when moving through a curved dip at a constant speed, calculated considering both gravitational and centripetal forces, is 60,000 N.
Explanation:
To calculate the normal force exerted by the road on the system (car and its occupants), we first note that when an object is in circular motion, the net force acting on the object is directed towards the center of the circle. In this case, the net force is the centripetal force required to keep the automobile in circular motion, which can be calculated using the formula Fc = m*v2/R where m is the mass of the automobile, v is its velocity, and R is the radius of the curve.
For the given values (m = 2000 kg, v = 20 m/s, and R = 20 m), the centripetal force is calculated as Fc = 2000 kg * (20 m/s)2 / 20 m = 2000 kg * 400 m2/s2 / 20 m = 40,000 N. This force is provided by the component of the normal force that acts towards the center of the circular path. Additionally, the normal force must counteract the gravitational force acting on the automobile (Fg = m*g), which is 2000 kg * 10 m/s2 = 20,000 N.
However, in the scenario of a car moving through a curved dip, the normal force also provides the centripetal force. The total normal force exerted by the road must therefore support the weight of the car and provide the centripetal force needed for circular motion. Thus, the total normal force is N = Fg + Fc = 20,000 N + 40,000 N = 60,000 N.
Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, such as sound and water waves. Suppose that your physics professor sets up two sound speakers in the front of your classroom and uses an electronic oscillator to produce sound waves of a single frequency. When she turns the oscillator on (take this to be its original setting), you and many students hear a loud tone while other students hear nothing.
The professor adjusts the oscillator to produce sound waves of twice the original frequency. What happens?
a. Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing.
b. The students who originally heard a loud tone again hear a loud tone, and the students who originally heard nothing still hear nothing.
c. Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone.
d. The students who originally heard a loud tone now hear nothing, and the students who originally heard nothing now hear a loud tone.
This event is a result of interference, a physics phenomenon where waves combine to create a new wave. Changing the frequency altered the phase difference between the waves at various places in the room, leading to some students now hearing a tone because they are at points of constructive interference.
Explanation:The correct answer is c. Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone. This situation is described by the phenomenon of interference. Interference occurs when two waves combine to form a resultant wave. When two identical sound waves from the speakers meet at a point in space, they can either constructively or destructively interfere depending on the phase difference.
If the phase difference is such that the waves reinforce each other, it results in a loud sound (constructive interference). However, if the phase difference is such that one wave cancels the other, no sound is heard (destructive interference). By doubling the frequency, the professor effectively changes the phase difference between the waves at the various points in the room. Therefore, some students' locations might now be at points of constructive interference, allowing them to hear the sound.
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The correct answer is option a. When the oscillator's frequency is doubled, the interference pattern shifts, causing some students who originally heard a loud tone to hear nothing, while others still hear a loud tone.
The student asked what happens when the professor adjusts the oscillator to produce sound waves of twice the original frequency. The correct answer is a) Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing.
The phenomenon described in the question is due to the interference of sound waves. Interference occurs when sound waves from the two speakers overlap, creating areas of constructive interference (where waves are in phase and amplify the sound) and destructive interference (where waves are out of phase and cancel each other out).
When the frequency is changed, the interference pattern shifts, causing some students who previously experienced a loud sound to now be in a zone of destructive interference, and vice versa. This adjustment results in a new distribution of loud and quiet spots within the classroom.
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
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
A student solving for the acceleration of an object has applied appropriate physics principles and obtained the expression a=a1 + F/m where a1 = 3.00 meter/ second2 F=12.0 kilogram.meter/second2 and m=7.00 kilogram. First which of the following is the correct step for obtaining a common denominator for the two fractions in the expression in solving for a?a. (m/m times a1/1) + (1/1 times F/m)b. (1/m times a1/1) + (1/m times F/m)c. (m/m times a1/1) + (F/F times F/m)d. (m/m times a1/1) +(m/m times F/m )
The correct step for obtaining a common denominator in the expression 'a = a1 + F/m' is to multiply each term by m/m, which results in '(m/m * a1/1) + (m/m * F/m)'. This leads to the acceleration, a, equalling 4.71 m/s2 when substituting the given values into the expression.
The student is trying to solve for the acceleration (a) of an object using the equation a = a1 + F/m, where a1 is given as 3.00 meters/second2, F as 12.0 kilogram.meter/second², and m as 7.00 kilograms. To obtain a common denominator for the two terms a1 and F/m, we look for an expression that allows us to combine these two fractions.
The correct step for obtaining a common denominator is option d. This is written as:
(m/m * a1/1) + (m/m * F/m)
This is because multiplying by m/m is equivalent to multiplying by 1, which does not change the value of the expression. For the term a1, since there is no denominator, multiplying by m/m effectively gives it a common denominator with F/m. The calculation becomes:
a = (m * a1 + F) / m
Substituting the given values:
a = (7.00 kg * 3.00 m/s2 + 12.0 kg*m/s2) / 7.00 kg
= (21.00 + 12.00) kg*m/s2 / 7.00 kg
= 33.00 kg*m/s2 / 7.00 kg
= 4.71 m/s2
Thus, the newton's second law, a = F/m, can be used to calculate the acceleration of the object.
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.
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
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]
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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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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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 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]
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]
PART 1/2
A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.
Assume: The mass of the atomic nucleus is
about 14.1 the mass of the neutron.
What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?
PART 2/2
If the initial kinetic energy of the neutron is
6.98 × 10−13 J, find its final kinetic energy.
Answer in units of J.
Answer:
0.247
5.25×10⁻¹³ J
Explanation:
Part 1/2
Elastic collision means both momentum and energy are conserved.
Momentum before = momentum after
m v = m v₁ + 14.1m v₂
v = v₁ + 14.1 v₂
Energy before = energy after
½ m v² = ½ m v₁² + ½ (14.1m) v₂²
v² = v₁² + 14.1 v₂²
We want to find the fraction of the neutron's kinetic energy is transferred to the atomic nucleus.
KE/KE = (½ (14.1m) v₂²) / (½ m v²)
KE/KE = 14.1 v₂² / v²
KE/KE = 14.1 (v₂ / v)²
We need to find the ratio v₂ / v. Solve for v₁ in the momentum equation and substitute into the energy equation.
v₁ = v − 14.1 v₂
v² = (v − 14.1 v₂)² + 14.1 v₂²
v² = v² − 28.2 v v₂ + 198.81 v₂² + 14.1 v₂²
0 = -28.2 v v₂ + 212.91 v₂²
0 = -28.2 v + 212.91 v₂
28.2 v = 212.91 v₂
v₂ / v = 28.2 / 212.91
v₂ / v = 0.132
Therefore, the fraction of the kinetic energy transferred is:
KE/KE = 14.1 (0.132)²
KE/KE = 0.247
Part 2/2
If a fraction of 0.247 of the initial kinetic energy is transferred to the atomic nucleus, the remaining 0.753 fraction must be in the neutron.
Therefore, the final kinetic energy is:
KE = 0.753 (6.98×10⁻¹³ J)
KE = 5.25×10⁻¹³ J
When a neutron collides elastically with the nucleus of an atom, a fraction of its kinetic energy is transferred to the nucleus. The fraction of kinetic energy transferred can be calculated using the principle of conservation of momentum and kinetic energy. For the given scenario, the fraction is 0.8636. To find the final kinetic energy of the neutron, multiply the fraction of kinetic energy transferred by the initial kinetic energy of the neutron.
Explanation:When a neutron in a reactor undergoes an elastic head-on collision with the nucleus of an atom initially at rest, kinetic energy is transferred from the neutron to the atomic nucleus. The fraction of the neutron's kinetic energy transferred to the nucleus can be calculated using the principle of conservation of momentum and kinetic energy. Since the mass of the atomic nucleus is about 14.1 times the mass of the neutron, the fraction of kinetic energy transferred can be calculated as:
Fraction of kinetic energy transferred = (14.1 - 1) / (14.1 + 1) = 0.8636
For PART 2/2, to find the final kinetic energy of the neutron, we can multiply the fraction of kinetic energy transferred to the nucleus by the initial kinetic energy of the neutron:
Final kinetic energy = Fraction of kinetic energy transferred x Initial kinetic energy = 0.8636 x 6.98 × 10-13 J
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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
If C is the curve given by r(t)=(1+2sint)i+(1+3sin2t)j+(1+1sin3t)k, 0≤t≤π2 and F is the radial vector field F(x,y,z)=xi+yj+zk, compute the work done by F on a particle moving along C.
The work done by vector field F on a particle moving along curve C can be computed using the line integral ∫F.dr. To compute the line integral, we need to parameterize curve C and evaluate the dot product of the vector field and the parameterized curve.
Explanation:To compute the work done by a vector field F on a particle moving along a curve C, we can use the line integral. The line integral of a vector field F along a curve C can be computed using the formula: ∫F.dr. In this case, F(x, y, z) = xi + yj + zk and C is given by r(t) = (1 + 2sin(t))i + (1 + 3sin(2t))j + (1 + sin(3t))k. We need to parameterize C to compute the line integral. Let's rewrite r(t) as:
r(t) = i + 2sin(t)i + j + 3sin(2t)j + k + sin(3t)k
We can then calculate the line integral using the given parameterization. Substituting r(t) into F(x, y, z), we get:
F(r(t)) = (1 + 2sin(t))i + (1 + 3sin(2t))j + (1 + sin(3t))k
Now, we can compute the line integral by evaluating ∫F(r(t)).dr over the given interval 0 ≤ t ≤ π/2. This involves evaluating the dot product of F(r(t)) and r'(t). The work done by F on the particle moving along C is the value of the line integral.
Final answer:
To compute the work done by a vector field on a particle moving along a curve, we can use the line integral formula. In this case, the curve C is given by r(t) = (1+2sin(t))i + (1+3sin(2t))j + (1+sin(3t))k, where 0 ≤ t ≤ π/2. The vector field F(x, y, z) = xi + yj + zk. The work done is equal to the line integral of F along C.
Explanation:
To compute the work done by a vector field on a particle moving along a curve, we can use the line integral formula:
Work = ∫C F · dr
In this case, the curve C is given by r(t) = (1+2sin(t))i + (1+3sin(2t))j + (1+sin(3t))k, where 0 ≤ t ≤ π/2. The vector field F(x, y, z) = xi + yj + zk.
The work done is equal to the line integral of F along C, so:
Work = ∫0^π/2 F(r(t)) · r'(t) dt
Now, we can substitute the given expressions for F and r(t) and evaluate the integral to find the work done.
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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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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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
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.
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.
Which term below best describes the forces on an object with a a net force of zero?
A. Inertia
B. Balanced Forces
C. Acceleration
D. Unbalanced Forces
Answer:
B. Balanced Forces
Explanation:
The net force is defined as the sum of all the forces acting on an object. Therefore, if the forces are balanced, they will counteract each other, causing the net force to be zero, then the object will continue at rest or moving with constant velocity.
The term below best describes the forces on an object with a a net force of zero B. Balanced Force
Which term below best describes the forces on an object with a a net force of zero?Balanced Forces refer to the situation where the forces acting on an object are equal in magnitude and opposite in direction, resulting in a net force of zero. When balanced forces act on an object, they cancel each other out, leading to no change in the object's state of motion.
This concept is tied to Newton's First Law of Motion, which states that an object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless acted upon by an unbalanced force.
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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.
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.
Arrange the listed objects according to their angular speeds, from largest to smallest.
a tire of radius 0.381 m rotating at 12.2 rpm
a bowling ball of radius 12.4 cm rotating at 0.456 rad/s
a top with a diameter of 5.09 cm spinning at 18.7∘ per second
a rock on a string being swung in a circle of radius 0.587 m with
a centripetal acceleration of 4.53 m/s2 a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s
Answer:
Explanation:
1 ) tire of radius 0.381 m rotating at 12.2 rpm
12.2 rpm = 12.2 /60 rps
n = .20333 rps
angular speed
= 2πn
= 2 x 3.14 x .20333
= 1.277 rad / s
2 ) a bowling ball of radius 12.4 cm rotating at 0.456 rad/s
angular speed = .456 rad/s
3 ) a top with a diameter of 5.09 cm spinning at 18.7∘ per second
18.7° per second = (18.7 / 180) x 3.14 rad/s
= .326 rad/s
4 )
a rock on a string being swung in a circle of radius 0.587 m with
a centripetal acceleration of 4.53 m/s2
centripetal acceleration = ω²R
ω is angular velocity and R is radius
4.53 = ω² x .587
ω = 2.78 rad / s
5 )a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s
The radius of the circle in which corner is moving
= .123 x √2
=.174 m
angular velocity = linear velocity / radius
.287 / .174
1.649 rad / s
The perfect order is
4 ) > 5> 1 >2>3.
Explanation:
To arrange the listed objects according to their angular speeds from largest to smallest, we must first calculate the angular speed ω (in radians per second) for each object, as this unit provides a common ground for comparison. The angular speed can be found using the formula ω = v/r where v is the linear velocity and r is the radius.
A tire rotating at 12.2 rpm: To convert this to radians per second, we use the conversion factor 1 rpm = 2π/60 rad/s. ω = 12.2 * (2π/60) = 1.279 rad/s.A bowling ball rotating at 0.456 rad/s: This is already in the correct unit, so ω = 0.456 rad/s.A top spinning at 18.7° per second: Converting degrees to radians (ω = 18.7 * (π/180)) gives ω = 0.326 rad/s.A rock with a centripetal acceleration of 4.53 m/s²: Using the formula a = ω² * r, we can rearrange to find ω (angular speed) = √(a/r) = √(4.53/0.587) = 2.780 rad/s.A square with corners moving at a tangential speed of 0.287 m/s: The radius for a square rotating about its center is half the diagonal length. For a square of 0.123 m side, the diagonal (d) = √(2) * side = √(2) * 0.123 m, and the radius (r) = d/2. So, ω = v/r = 0.287/(√(2)*0.123/2) = 4.205 rad/s.Arranging these angular speeds from largest to smallest:
Angular speed of square: 4.205 rad/sAngular speed of rock: 2.780 rad/sAngular speed of tire: 1.279 rad/sAngular speed of bowling ball: 0.456 rad/sAngular speed of top: 0.326 rad/sA 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
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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