Answer:
Explanation:
Heat energy given out by the soup
= C_v x ( t₂ - t₁ )
= 33 x ( 340 - 300)
= 1320 J
This heat is given to Carnot engine . Efficiency of engine
= (340 - 300 ) / 340
= 40 / 340
2 / 17
This fraction of total heat given is converted into useable work by the engine.
You throw a ball of mass 1 kg straight up. You observe that it takes 2.2 s to go up and down, returning to your hand. Assuming we can neglect air resistance, the time it takes to go up to the top is half the total time, 1.1 s. Note that at the top the momentum is momentarily zero, as it changes from heading upward to heading downward.
(a) Use the momentum principle to determine the speed that the ball had just AFTER it left your hand.
vinitial = ?? m/s
(b) Use the Energy Principle to determine the maximum height above your hand reached by the ball.
h = ?? m
Answer:
10.791 m/s
5.93505 m
Explanation:
m = Mass of ball
[tex]v_f[/tex] = Final velocity
[tex]v_i[/tex] = Initial velocity
[tex]t_f[/tex] = Final time
[tex]t_i[/tex] = Initial time
g = Acceleration due to gravity = 9.81 m/s²
From the momentum principle we have
[tex]\Delta P=F\Delta t[/tex]
Force
[tex]F=mg[/tex]
So,
[tex]m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s[/tex]
The speed that the ball had just after it left the hand is 10.791 m/s
As the energy of the system is conserved
[tex]K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m[/tex]
The maximum height above your hand reached by the ball is 5.93505 m
The speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.
What is momentum principal?When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the principal of momentum.
The net force using the momentum principle can be given as,
[tex]F_{net}=\dfrac{\Delta P}{\Delta t}[/tex]
Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. Therefore, the above equation for initial and final velocity and time can be written as,
[tex]F_{net}=\dfrac{m(v_f-v_i)}{(t_f-t_i)}[/tex]
(a) The speed that the ball had just AFTER it left your hand-The mass of the ball is 1 kg, and it takes 2.2 s to go up and down. The net force acting on the body is mass times gravitational force.
Therefore, the above equation can be written as,
[tex]mg=\dfrac{m(v_f-v_i)}{(t_f-t_i)}\\g=\dfrac{(v_f-v_i)}{(t_f-t_i)}\\[/tex]
As the final velocity is zero and initial time is also zero. Therefore,
[tex](-9.81)=\dfrac{(0-v_i)}{(1.1-0)}\\v_i=10.791 \rm m/s[/tex]
(b) The maximum height above your hand reached by the ball-Using the energy principle, we can equate the kinetic energy of the system to the potential energy of the system as,
[tex]\dfrac{1}{2}mv_i^2=mgh\\\dfrac{1}{2}(10.791)^2=(9.81)h\\h=5.94\rm m[/tex]
Thus, the speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.
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A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal.
At its highest point in this trajectory, the velocityof the projectile would be what?
Answer:
169.4 m/s
Explanation:
Given that the angle of projectile is θ_1 =450°
The speed of body at maximum height is U cosθ_1 = 150 m/s
The angle in second trail is θ_2 =37°
From the given data U cosθ_1 = 150 m/s
U = 150 m/s / cosθ_1
= 150m/s / cos45°
=212.13 m/s
The velocity of the projectile at maximum height in second trail= Ucos(θ_2)
=212.13 m/s×cos37°
=169.4 m/s
The velocity of a projectile at its highest point when fired at an angle to the horizontal remains the same if the initial speed is unchanged, regardless of the angle. Hence, even when changing the angle from 45 to 37 degrees, the velocity at the highest point would still be 150 m/s.
The student's question deals with the velocity of a projectile at the highest point in its trajectory. When a projectile is fired upward at an angle, its velocity at the highest point of its trajectory is only composed of the horizontal component because the vertical component of the velocity becomes zero at that point.
In the given scenario, when the projectile is shot at 45 degrees to the horizontal, the speed at the highest point is given as 150 m/s. This speed represents the horizontal component since there's no vertical component at the highest point. When the angle is changed to 37 degrees, the horizontal component of the initial velocity is calculated using the cosine component of the initial speed, hence the velocity at the highest point remains the same as when it was fired at 45 degrees, provided that the initial speed is unchanged.
Therefore, in the second trial, with the angle at 37 degrees, the velocity of the projectile at its highest point would still be 150 m/s, because the horizontal component of the velocity is not affected by the change in angle.
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s2, 2.7 m, 6 kg, 31C. What is the speed of the ball when it is in circular motion? Answer in units of m/s.
The speed of the metal ball in circular motion is calculated by using the equation for centripetal acceleration (v = √(rg)) and substituting the given values for radius and gravity. After calculation, the speed of the ball is determined to be approximately 5.1 m/s.
Explanation:In this physics problem, we can calculate the speed of the metal ball by using the provided values and the concept of centripetal acceleration. Centripetal acceleration results from the force that makes the metal ball move in a circular pattern, which in this case is provided by the tension in the thread and gravity. In a scenario where the ball is in uniform circular motion, with the tension at an angle, forming a cone around the vertical axis, the component of tension providing the centripetal force equals the gravitational force or weight of the object.
A key equation that can be used here is: ac = v² /r, where 'ac' is the centripetal acceleration, 'v' is the velocity (or speed), and 'r' is the radius. Given that ac = g (acceleration due to gravity = 9.8 m/s²), we can modify the equation to: v = √(rg), where 'r' is the radius or the length of the thread (2.7 m).
Implementing these values into our equation we find that: v = √[(2.7 m)(9.8 m/s²)] = √26.46 = 5.1 m/s. Therefore, the speed of the metal ball when it is in circular motion is approximately 5.1 m/s.
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A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point where the spring is not stretched. Then the mass is released and begins to oscillate. The lowest point in the oscillation is 18 cm below the point where the mass was released. What is the oscillation frequency?
Answer:
The oscillation frequency of the spring is 1.66 Hz.
Explanation:
Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :
[tex]E_s=\dfrac{1}{2}kx^2[/tex]
k is the spring constant
x is the distance to the lowest point form the initial position.
When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.
[tex]E_P=mgx[/tex]
Equating these two energies,
[tex]\dfrac{1}{2}kx^2=mgx[/tex]
[tex]\dfrac{k}{m}=\dfrac{2g}{x}[/tex].............(1)
The expression for the oscillation frequency is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}}[/tex] (from equation (1))
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}[/tex]
f = 1.66 Hz
So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.
Point A has coordinates of (2,5,1) and point D has coordinate of (9, 6, 10). Calculate the unit vector of (landa AD)
Answer:(69,69,69)
Explanation:
The ratio of the intercortex of the inner circle has an area of 98:89. This is due to the fact the vector of the circle is no less than the ratio times the pie squared. Then pie squared the circumference will end up approximately at 68.83957 where you then round up to all 69's.
What two elements are used with other metals to lower their melting points?
A) antimony and carbon
B) bismuth and carbon
C) antimony and bismuth
D) arsenic and antimony
Answer:
C) antimony and bismuth
Explanation:
Answer:
its C
Explanation:
Why would a plume of solid silicate rock rising slowly from deep in the mantle begin melting as it neared the base of the lithosphere?
A. The rock heats up and expands at lower pressures, causing it to liquefy
B. Temperatures remain high as lowered pressures decrease melting temperatures.
C. The lowered pressures cause rapid heat loss accompanied by melting.
D. None of the above
Answer:
B. Temperatures remain high as lowered pressures decrease melting temperatures.
Explanation:
Rocks , like ice , contracts on melting so , their melting point decreases on decrease in pressure . When hot solid rocks are pushed towards the surface of the earth, due to decrease in pressure , their melting point decreases , so the solid rocks starts melting without any addition of heat . Their present temperature becomes more than their melting point.
An automatic coffee maker uses a resistive heating element to boil the 2.4 kg of water that was poured into it at 21 °C. The current delivered to the coffee pot is 8.5 A when it is plugged into a 120 V electrical outlet. If the specific heat capacity of water is 4186 J/kgC° , approximately how long does it take to boil all of the water?
Answer:
Explanation:
The expression for the calculation of the enthalpy change of a process is shown below as:-
[tex]\Delta H=m\times C\times \Delta T[/tex]
Where,
[tex]\Delta H[/tex]
is the enthalpy change
m is the mass
C is the specific heat capacity
[tex]\Delta T[/tex]
is the temperature change
Thus, given that:-
Mass of water = 2.4 kg
Specific heat = 4.18 J/g°C
[tex]\Delta T=100-21\ ^0C=79\ ^0C[/tex]
So,
[tex]\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ[/tex]
Heat Supplied [tex]Q=VIt[/tex]
where [tex]i=current[/tex]
[tex]V=Voltage[/tex]
[tex]8.5\times 120\times t=2.4\times 4186\times 79[/tex]
[tex]t=778.10 s[/tex]
Final answer:
To determine the time required to boil the water in a coffee maker, calculate the power using P = IV, then the heat needed to raise the water temperature to its boiling point with Q = mcΔT, and finally divide Q by P to find the time.
Explanation:
The question deals with calculating how long it takes an automatic coffee maker to boil 2.4 kg of water using a resistive heating element. The initial temperature of the water is 21 °C, and we are given that the coffee maker is plugged into a 120 V electrical outlet with a current of 8.5 A. The specific heat capacity of water is 4186 J/kg°C. To find the time to boil the water, we'll first calculate the power delivered by the heating element using the relationship P = IV, where P is power, I is current, and V is voltage. Once power is known, we can use the equation Q = mcΔT to find the heat Q required to raise the temperature of water to its boiling point (100 °C), where m is water's mass, c is the specific heat capacity, and ΔT is the change in temperature. Finally, we'll divide quantified heat Q by the power P to find the time t: t = Q/P.
A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximately 281 micrometers on each side. Up close, you can see the individual pixels, but from a distance they appear to blend together and form the image on the screen.
If the maximum distance between the screen and your eyes at which you can just barely resolve two adjacent pixels is 1.30 meters, what is the effective diameter d of your pupil? Assume that the resolvability is diffraction-limited. Furthermore, use lambda = 550 nanometers as a characteristic optical wavelength.
The effective diameter d of your pupil should be considered as the 0.0031 when maximum distance should be 1.30 meters.
Calculation of the effective diameter:Since
y = Length of pixel = 281 μm
L = Distance to screen = 1.3 m
Wavelength = 550 nm
d = Pupil diameter
So,
[tex]tan = y / l\\\\= tan^-1 * 281*10^-6/1.3[/tex]
Now
sin = 1.22 wavelength / d
d = 1.22 * wavelength / sin
[tex]= 1.22*550*10^-9/ sin ( tan^-1 * 281*10^-6/1.3)[/tex]
= 0.0031 m
hence, The effective diameter d of your pupil should be considered as the 0.0031 when maximum distance should be 1.30 meters.
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The effective diameter of the pupil of one's eye when barely able to resolve two adjacent pixels on a computer screen from a certain distance, where the point of resolution is diffraction limited, is approximately 1.7 mm.
Explanation:This question involves the use of the formula for the resolution of an optical system, which is the Rayleigh’s criterion. The formula can be written as follows: θ = 1.22 * λ/D, where θ is the smallest angular separation that can be resolved, λ is the wavelength of light, and D is the diameter of the aperture (i.e., the pupil in this case).
In this question, the pixel’s size and the distance from the screen allow us to calculate θ. The tangent of θ is approximately equal to the ratio of the pixel size to the viewing distance. Hence, θ = arctan((281 * 10^-6 m) / 1.3 m).
By substituting θ, and λ (550 * 10^-9 m) into the formula and solve for D, the effective diameter of the pupil, we get: D = 1.22 * λ / θ. The result is around 0.0017 meters or 1.7 mm.
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To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0°.What is the angular acceleration (in rad/s2) if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.250 kg · m2, and the net force she exerts is 550 N at an effective perpendicular lever arm of 2.10 cm? (Ignore gravity)How much work (in J) does she do?
To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:
[tex]\tau = Fr[/tex]
Where,
F = Force
r = Radius
Replacing we have that,
[tex]\tau = Fr[/tex]
[tex]\tau = 21cm (\frac{1m}{100cm})* 550N[/tex]
[tex]\tau = 11.55Nm[/tex]
The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore
[tex]I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2[/tex]
[tex]I = 0.394kg\cdot m^2[/tex]
Finally, angular acceleration is a result of the expression of torque by inertia, therefore
[tex]\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}[/tex]
[tex]\alpha = \frac{11.55}{0.394}[/tex]
[tex]\alpha = 29.3 rad/s^2[/tex]
PART B)
The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians [tex](\pi / 3)[/tex], therefore
[tex]W = \tau \theta[/tex]
[tex]W = 11.5* \frac{\pi}{3}[/tex]
[tex]W = 12.09J[/tex]
A small object is attached to the end of a relaxed, horizontal spring whose opposite end is fixed. The spring rests on a frictionless surface. Let the initial position of the object be defined as x-0. The object is pulled to position x = A and then released, after which it undergoes simple harmonic motion. In one full cycle of its motion, the total distance traveled by the object is 2 4 O 4A
Answer:
The total distance traveled by an object attached to a spring that is pulled to position x=A and then released is 4 A.
Explanation:
We have an small object attached to a relaxed spring whose opposite end is fixed. The spring rests on a frictionless surface. This means the only force acting on the object is the elastic force of the spring, a conservative force, since the weight and the normal force compensate between them.
The initial position of the object is x=0 then is pulled to position x=A and released. After which it undergoes a simple harmonic motion with an amplitude A. From the position x=A to the equilibrium position in x=0 the object travels a distance A. From the equilibrium position x=0 to maximum negative position in x= -A the object travels again a distance A. Then to return to the original position the object should travel a distance 2 A in reverse direction.
In one full cycle of its motion the object travels a distance 4 A.
How long does it take a 750-W coffeepot to bring to a boil 0.85L of water initially at 15?C?
Assume that the part of the pot which is heated with the water is made of 360 g of aluminum, and that no water boils away. Ignore the heat loss to the surrounding environment.
The value of specific heat for water is 4186 J/kg?C? and for aluminum is 900 J/kg?C?.
To calculate the time it takes for the coffee pot to bring the water to a boil, we need to determine the amount of heat required and use the power of the coffee pot. By calculating the heat required for both the water and the aluminum pot, and using the power of the coffee pot, we can determine that it takes approximately 96.077 seconds for the coffee pot to bring the water to a boil.
Explanation:To determine how long it takes for the coffee pot to bring the water to a boil, we need to calculate the amount of heat required. The heat required to raise the temperature of the water from 15?C to 100?C is given by the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature. Using the values given, we can calculate the heat required.
The heat required for the water is Q = (0.85 kg)(4186 J/kg?C)(100?C - 15?C) = 43707.5 J. The heat required for the aluminum coffee pot is Q = (0.36 kg)(900 J/kg?C)(100?C - 15?C) = 28350 J. Adding the two heats together gives us a total heat of 72057.5 J.
The power of the coffee pot is given as 750 W. Power is defined as the rate at which work is done or the rate at which energy is transferred. So, we can use the formula P = E/t, where P is power, E is energy, and t is time. Rearranging the formula, we can solve for t:
t = E/P = 72057.5 J / 750 W = 96.077 seconds.
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Question:
Jane, looking for Tarzan, is running at top speed (4.5 m/s) and grabs a vine hanging vertically from a tall tree in the jungle.
How high can she swing upward?
Energy Conservation
According to the conservation of the mechanical energy the total energy remains constant of the system, therefore the kinetic energy of the system will get converted into the potential energy of the system.
Answer:
She can swing 1.0 m high.
Explanation:
Hi there!
The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).
The kinetic energy is calculated as follows:
KE = 1/2 · m · v²
And the potential energy:
PE = m · g · h
Where:
m = mass of Jane.
v = velocity.
g = acceleration due to gravity (9.8 m/s²).
h = height.
Then:
ME = KE + PE
Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:
ME = KE + PE (PE = 0)
ME = KE
ME = 1/2 · m · (4.5 m/s)²
ME = m · 10.125 m²/s²
When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:
ME = KE + PE (KE = 0)
ME = PE
ME = m · 9.8 m/s² · h
Then, equallizing both expressions of ME and solving for h:
m · 10.125 m²/s² = m · 9.8 m/s² · h
10.125 m²/s² / 9.8 m/s² = h
h = 1.0 m
She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).
17 of 24 Constants In an experiment to simulate conditions within an automobile engine, 0.165 mol of air at a temperature of 700 K and a pressure of 3.40×106 Pa is contained in a cylinder of volume 280 cm3 . Then 690 J of heat is transferred to the cylinder. Part APart complete If the volume of the cylinder is constant while the heat is added, what is the final temperature of the air? Assume that the air is essentially nitrogen gas. T = 901 K Previous Answers Correct Part B If instead the volume of the cylinder is allowed to increase while the pressure remains constant, find the final temperature of the air.
Answer:
a. T₂ = 901.43 ° K
b. T₂ = 843.85 ° K
Explanation:
Given
η = 0.165 mol , Q = 690 J , V = 280 cm³ , P = 3.40 x 10 ⁶ Pa , T₁ = 700 ° K
Using the equation that describe the experiment of heat
Q = η * Cv * ΔT
a.
Nitrogen Cv = 20.76 J / mol ° K
ΔT = T₂ - T₁
T₂ = [ Q / ( η * Cv) ] + T₁
T₂ = [ 690 J / ( 0.165 mol * 20.76 J / mol ° K ) ] + 700 ° K
T₂ = 901.43 ° K
b.
Nitrogen Cp = 29.07 J / mol ° K
T₂ = [ Q / ( η * Cv) ] + T₁
T₂ = [ 690 J / ( 0.165 mol * 29.07 J / mol ° K ) ] + 700 ° K
T₂ = 843.85 ° K
The Final temperature of the air when the volume of the cylinder is constant is 700 K.
To find the final temperature of the air when the volume of the cylinder is constant, we can use the ideal gas law equation:
P₁V₁/T₁ = P₂V₂/T₂
Since the volume is constant, V₁ = V₂.
The initial pressure (P₁) and temperature (T₁) are given, and the final pressure (P₂) is also given (3.40×106 Pa).
We can rearrange the equation to solve for T₂:
T₂ = (P₂T₁) / P₁
Substituting the given values, we have:
T₂ = (3.40×106 Pa * 700 K) / (3.40×106 Pa)
T₂ = 700 K
So, the final temperature of the air is 700 K when the volume of the cylinder is constant.
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The diagram shows models of both prokaryotic and eukaryotic cells. The diagram shows models of both prokaryotic and eukaryotic cells. Prokaryotic cells are considered to be the more primitive of the two. What is a similarity between eukaryotes and prokaryotes as illustrated by the models? A) Both contain ribosomes. B) Both contain mitochondria. C) Both are surrounded by a capsule. D) Both contain a membrane-bound nucleus.
Answer:
both have plasma membrane, ribosomes, cytoplasm, and DNA
Explanation:
Both eukaryotic and prokaryotic cells contain ribosomes, which are responsible for protein synthesis. Eukaryotic cells are more complex with a true nucleus, found in animals and plants, while prokaryotic cells, found in bacteria, are simpler and lack a true nucleus.
Explanation:Option A is the correct answer. Both eukaryotic and prokaryotic cells contain ribosomes. The ribosomes in both types of cells serve the same function: protein synthesis, or assembly of amino acids to form proteins. Eukaryotic cells are those with a true nucleus and complex structures, found in animals, plants, fungi, and protists, while prokaryotic cells are smaller, simpler cells without a true nucleus, found in bacteria and archaea.
Unlike eukaryotic cells, prokaryotic cells don't contain a membrane-bound nucleus or mitochondria. A capsule is found in some bacteria, not all prokaryotes and no eukaryotes.
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As part of a safety investigation, two 1600 kg cars traveling at 21 m/s are crashed into different barriers.
A) Find the average force exerted on the car that hits a line of water barrels and takes 1.8 s to stop.
B) Find the average force exerted on the car that hits a concrete barrier and takes 0.14 s to stop.
Answer
given,
mass of the car = 1600 Kg
velocity of the car = 21 m/s
time = 1.8 s
a) momentum = mass x velocity
P = 1600 x 21
P = 33600 kg.m/s
[tex]Force = \dfrac{change\ in \ momentum}{t}[/tex]
[tex]Force = \dfrac{33600-0}{1.8}[/tex]
F = 18666.67 N
b) [tex]Force = \dfrac{change\ in \ momentum}{t}[/tex]
[tex]Force = \dfrac{33600-0}{0.14}[/tex]
F = 240000 N
The two cars collide at right angles in the intersection of two icy roads. Car A has a mass of 1965 kg and car B has a mass of 1245 kg. The cars become entangled and move off together with a common velocity in the direction indicated. If car A was traveling 52 km/h at the instant of impact, compute the corresponding velocity of car B just before impact.
Answer:
U2 = 47.38m/s = initial velocity of B before impact
Explanation:
An example of the diagram is shown in the attached file because of missing angle of direction in the question
Mass A, B are mass of cars
A = 1965
B =1245
U1 = initial velocity of A = 52km/hr
U2 = initial velocity of B
V = common final velocity of two cars
BU2 = (A + B)*V sin ¤ ...eq1 y plane
AU1 = (A + B) *V cos ¤ ....equ 2plane
From equ 2
V = AU1/(A + B)*cos ¤
Substitute V into equation 1
We have
U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°
Substitute all parameters to get
U2 = (1965/1245)*52 * tan 30°
U2 = 47.38m/s
To find the corresponding velocity of car B just before impact, we can use the principle of conservation of momentum. By equating the total momentum before the collision to the total momentum after the collision, we can solve for the velocity of car B. Plugging in the given values and calculating, we can find the corresponding velocity of car B just before impact.
Explanation:To solve this problem, we can use the principle of conservation of momentum. Since the two cars stick together after the collision, the total momentum before the collision is equal to the total momentum after the collision.
Let's define the velocity of car A just before impact as vA and the velocity of car B just before impact as vB.
Using the principle of conservation of momentum, we can write:
(mass of car A * velocity of car A) + (mass of car B * velocity of car B) = (total mass * common velocity)
Plugging in the given values, we have:
(1965 kg * 52 km/h) + (1245 kg * vB) = (3210 kg * common velocity) (Note: Convert km/h to m/s)
Simplifying the equation, we get:
101820 kg * km/h + 1245 kg * vB = 3210 kg * common velocity
Now, we can solve for the velocity of car B just before impact:
vB = (3210 kg * common velocity - 101820 kg * km/h) / 1245 kg
By substituting the values and calculating, we can find the corresponding velocity of car B just before impact.
The hydrofoil boat has an A-36 steel propeller shaft that is 100ft long. It is connected to an inline diesel engine that delivers a maximum power of 2500 hp and causes the shaft to rotate at 1700 rpm. If the outer diamater of the shaft is 8 in. and the wall thickness is 3/8 in., determine the maximum sheer stress developed in the shaft. Also, what is the "wind up," or angle of twist in the shaft at full power?
Final answer:
The maximum shear stress developed in the shaft is approximately 7.57 MPa. The angle of twist in the shaft at full power is approximately 3.00°.
Explanation:
To determine the maximum shear stress developed in the shaft, we can use the formula:
Shear stress (τ) = Torque (T) / Polar Moment of Inertia (J)
To find the torque, we can use the formula:
Torque (T) = Power (P) / Angular Velocity (ω)
We are given that the power is 2500 hp and the angular velocity is 1700 rpm. Converting these values to W and rad/s respectively, we have:
Power (P) = 2500 hp * 745.7 W/hp ≈ 1,864,250 W
Angular Velocity (ω) = 1700 rpm * 2π rad/minute ≈ 17897.37 rad/s
Substituting these values into the torque formula, we have:
Torque (T) = 1,864,250 W / 17897.37 rad/s ≈ 103.98 N*m
Next, we need to find the polar moment of inertia (J) of the shaft. The polar moment of inertia for a solid shaft can be calculated using the formula:
J = (π/2) * (outer diameter^4 - inner diameter^4)
Converting the diameter to meters, we have:
Diameter (d) = 8 in * 0.0254 m/in = 0.2032 m
Substituting this value into the polar moment of inertia formula, we have:
J = (π/2) * (0.2032^4 - (0.2032 - 2 * 3/8)^4)
= (π/2) * (0.2032^4 - 0.1926^4)
0.013743 m^4
Finally, we can calculate the maximum shear stress using the formula:
Shear stress (τ) = 103.98 N*m / 0.013743 m^4 ≈ 7.57 MPa
As for the angle of twist, we can use the formula:
Angle of twist (θ) = T * L / (G * Given that the length of the shaft is 100 ft and Young's modulus (G) for A-36 steel is approximately 200 GPa, we can calculate the angle of twist as follows:
Angle of twist (θ) = 103.98 N*m * 100 ft * 0.3048 m/ft / (200 GPa * 0.013743 m^4)
≈ 0.0524 radians or 3.00°
A projectile of mass m1 moving with speed v1 in the +x direction strikes a stationary target of mass m2 head-on. The collision is elastic. Use the Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target, making no approximations concerning the masses. (Use the following as necessary: m1, m2, and v1.)
Using Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target are respectively;
v1 = (m1•u1 - m2•u1)/(m1 + m2)
v1 = (m1•u1 - m2•u1)/(m1 + m2)v2 = (m1•u1 + m1•u2)/(m1 + m2)
Let us denote the following;
Mass of projectile; m1
Mass of target; m2
Initial velocity of projectile; u1
Final velocity of projectile; v1
Initial velocity of target; u2 = 0 m/s
Final velocity of target; v2
From conservation of momentum, we have;
(m1•u1) + (m2•u2) = (m1•v1) + (m2•v2)
Plugging in the relevant values;
(m1•u1) + (m2•0) = (m1•v1) + (m2•v2)
(m1•u1) = (m1•v1) + (m2•v2) - - - (eq 1)
From energy principle, we have;
u1 + v1 = u2 + v2
Thus;
Since u2 is zero, then we have;
u1 + v1 = v2 - - - (eq 2)
Put (u1 + v1) for v2 in eq 1 to get;
(m1•u1) = (m1•v1) + (m2(u1 + v1))
(m1•u1) = (m1•v1) + (m2•u1 + m2•v1) - - - (eq 3)
Let us make v1 the subject;
m1•u1 - m2•u1 = m1•v1 + m2•v1
v1(m1 + m2) = m1•u1 - m2•u1
v1 = (m1•u1 - m2•u1)/(m1 + m2)
Let us make v1 the subject in eq 2;
v1 = v2 - u2
Put v2 - u2 for v1 in eq 1 to get;
(m1•u1) = (m1(v2 - u2) + (m2•v2)
(m1•u1) = m1•v2 - m1•u2 + m2•v2
m1•u1 + m1•u2 = v2(m1 + m2)
v2 = (m1•u1 + m1•u2)/(m1 + m2)
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The final velocities of the projectile and the target after an elastic collision can be arrived at by solving the equations for Conservation of Momentum and Conservation of Kinetic Energy simultaneously. Here, the target's initial velocity is zero (it's at rest). By substituting this into the equations and solving them, we can find the final velocities.
Explanation:An elastic collision is a type of collision where both momentum and kinetic energy are conserved. For this problem, we can solve the final velocities using the conservation laws.
Conservation of Momentum gives us:
m1*v1 = m1*v1' + m2*v2' -------- (1)
V1' and V2' are the final velocities of the projectile and target, respectively.
Conservation of Kinetic Energy gives us:
1/2*m1*v1^2 = 1/2*m1*(v1')^2 + 1/2*m2*(v2')^2 -------- (2)
While solving these two equations, one should remember that m2 was initially at rest, so the momentum in the Y-direction is zero. The above-mentioned equations can be solved by substitution method or any other algebraic method to get the final velocities v1' and v2'.
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A cylinder is filled with 10.0 L of gas and a piston is put into it. The initial pressure of the gas is measured to be 163. kPa. The piston is now pushed down, compressing the gas, until the gas has a final volume of 4.40 L. Calculate the final pressure of the gas. Round your answer to 3 significant digits.
Answer:
P₂ = 370 kPa
Explanation:
Boyle's law: States that the volume of a given mass of gas is inversely proportional to its pressure provided the temperature remains constant. It can be expressed mathematically as,
P₁V₁ = P₂V₂................ Equation 1
Where P₁ = Initial Pressure, V₁ = Initial volume, P₂ = Final Pressure, V₂ = Final Volume.
Making P₂ The subject of the equation above,
P₂ = P₁V₁/V₂..................... equation 2
Substituting these vaues into equation 2,
Where P₁ = 163 kPa = 163000 pa, V₁ = 10 L, V₂ = 4.4 L.
P₂ = 163000(10)/4.4
P₂ = 370454.55 Pa
P₂ = 370000 Pa
P₂ = 370 kPa To 3 significant figure.
Therefore Final pressure = 370 kPa
Which of the following best describes a substance in which the temperature remains constant while at the same time it is experiencing an inward heat flow?
a. gasb. liquidc. solidd. substance undergoing a change of state
Answer:
The right option is (d) substance undergoing a change of state
Explanation:
Latent Heat: Latent heat is the heat required to change the state of a substance without change in temperature. Latent heat is also known as hidden heat because the heat is not visible. The unit is Joules (J).
Latent heat is divided into two:
⇒ Latent Heat of fusion
⇒ Latent Heat of vaporization.
Latent Heat of fusion: This is the heat energy required to convert a substance from its solid form to its liquid form without change in temperature. E.g (Ice) When ice is heated, its temperature rise steadily until a certain temperature is reached when the solid begins to melts.
Latent Heat of vaporization: This is the heat required to change a liquid substance to vapor without a change in temperature. The latent heat depend on the mass of the liquid and the nature of the liquid. E.g When water is heated from a known temperature its boiling point (100°C) When more heat is supplied to its boiling temperature, it continue to boil without a change in temperature.
From The above, Latent heat brings about a change of state of a substance at a steady temperature.
The right option is (d) substance undergoing a change of state
The substance that best represents a temperature remaining constant despite an inward heat flow is one that is undergoing a change of state, such as during the melting of ice or boiling of water. So, the correct option is D.
The best description of a substance in which the temperature remains constant while it is experiencing an inward heat flow is a substance undergoing a change of state (option d). This occurs during a process called phase transition.
Here's an example: when ice melts into water, it absorbs heat from the surroundings but the temperature stays constant at 0 degrees Celsius until all of the ice has melted.
Similarly, when water is boiling, its temperature holds steady at 100 degrees Celsius until all the water has transitioned into steam, despite the continued input of heat.
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What temperature increase is necessary to increase the power radiated from an object by a factor of 8?
a. 8 Kb. 2 Kc. 100%d. about 68%
To solve this problem it is necessary to apply the concepts related to the Power defined from the Stefan-Boltzmann equations.
The power can be determined as:
[tex]P = \sigma T^4[/tex]
Making the relationship for two states we have to
[tex]\frac{P_1}{P_2} = \frac{T_1^4}{T_2^4}[/tex]
Since the final power is 8 times the initial power then
[tex]P_2 = 8P_1[/tex]
Substituting,
[tex]\frac{1}{8} = \frac{T_1^4}{T_2^4}[/tex]
[tex]T_2 = T_1 8*(\frac{1}{4})[/tex]
[tex]T_2 = 1,68T_1[/tex]
The temperature increase would then be subject to
[tex]\Delta T = T_2-T_1[/tex]
[tex]\Delta T = 1.68T_1 -T_1[/tex]
[tex]\Delta T = 0.68T_1[/tex]
The correct option is D, about 68%
Final answer:
The correct option is (c) 100%. To increase the power radiated from an object by a factor of 8, the temperature must be doubled, corresponding to a 100% increase. This is based on the Stefan-Boltzmann law.
Explanation:
The question relates to the physics concept of radiation and how it varies with temperature. Specifically, it involves the Stefan-Boltzmann law, which states that the power radiated from an object is proportional to the fourth power of its absolute temperature.
If the power radiated from an object is to increase by a factor of 8, we use the relationship P ≈ T^4, where P is power and T is the temperature in Kelvin. To calculate the necessary temperature increase, if the initial power is P and the final power is 8P, then (T_final)^4 = 8 × (T_initial)^4.
By taking the fourth root on both sides, we find that T_final = 2× T_initial. Therefore, the temperature must be doubled, or the temperature increase is by 100%. Hence, the correct answer is option (c) 100%
The process of vision involves 4 basic steps in an appropriate order. Of the options listed below, what is correct order in the process of vision? a) gathering light, within the eye processing, transduction, and brain processing b) without the eye processing, gather light, transfusion, and brain processing c) photo-production, gathering light, within the eye processing, and transduction d) gathering light, photo-production, transfusion, and within the eye processing e) light collection, photo-production, hoarding light, and brain processing
Answer:
Option (a) is correct.
Explanation:
For us to be able to visualize an object, there are basic processes that takes place in a split second.
First Light which strike the object reflect into our eye. This reflected light passes through the pupil as the iris gives way for it. This light moves through other layers and impinges on the retina and stimulates it. This is the process of within the eye processing. The light is then converted to a signal by the optic nerve and sent to the brain for interpretation. The image produced by the retina is inverted. The brain helps to interpret this image as upright. This is the reason why we see objects just the way they are
All Houston Methodist buildings system wide have an emergency power generator that turns on to supply emergency power after normal power shuts down within:
A. 5 seconds
B. 30 seconds
C. 60 seconds
D. 10 seconds
Answer:
(D) 10 seconds
Explanation:
The Houston Methodist Hospital automated the emergency power supply system (EPSS) testing processes in order to ensure the safety of patients during an outage
There is an emergency power generator that turns on to supply emergency power after normal power shuts down within 10 seconds.
Answer:
D. 10 seconds
Explanation:
The Houston Methodist Hospital has eleven diesen powered generators that turn on at most 10 seconds after the normal power shuts down.
Considering that it is a hospital, this system is really important.
So the correct answer is:
D. 10 seconds
A backyard swimming pool with a circular base of diameter 6.00 m is filled to depth 1.50 m. (a) Find the absolute pres- sure at the bottom of the pool. (b) Two persons with com- bined mass 150 kg enter the pool and float quietly there. No water overflows. Find the pressure increase at the bottom of the pool after they enter the pool and float.
Answer
given,
diameter = 6 m
depth = h = 1.5 m
Atmospheric pressure = P₀ = 10⁵ Pa
a) absolute pressure
P = P₀ + ρ g h
P = 10⁵ + 1000 x 10 x 1.5
P = 1.15 x 10⁵ Pa
b) When two person enters into the pool
mass of the two person = 150 Kg
weight of water level displaced is equal to the weight of person
ρ V g = 2 m g
[tex]V = \dfrac{2 m}{\rho}[/tex]
[tex]V = \dfrac{2\times 150}{1000}[/tex]
V = 0.3 m³
Area of pool = [tex]\dfrac{\pi}{4}d^2[/tex]
=[tex]\dfrac{\pi}{4}\times 6^2[/tex]
= 28.27 m²
height of the water rise
[tex]h = \dfrac{V}{A}[/tex]
[tex]h = \dfrac{0.3}{28.27}[/tex]
h = 0.0106 m
pressure increased
P = ρ g h
P = 1000 x 10 x 0.0106
P = 106.103 Pa
a. the absolute pressure at the bottom of the pool is approximately 116.02 kPa.
b. the pressure increase at the bottom of the pool after the persons enter and float is approximately 0.147 kPa.
To find the absolute pressure at the bottom of the pool, we can use the hydrostatic pressure formula:
(a) Absolute pressure at a depth in a fluid:
P=P_0 +ρ⋅g⋅h
Where:
P is the absolute pressure at the given depth.
P_0 is the atmospheric pressure (which we'll assume to be 1 atm, or approximately 101.3 kPa).
ρ is the density of the fluid (water), which is about 1000 kg/m³.
g is the acceleration due to gravity, approximately 9.81 m/s².
h is the depth below the surface.
Given that the diameter of the pool is 6.00 m, the radius (r) is 3.00 m, and the depth (h) is 1.50 m, we can find the pressure at the bottom of the pool:
P=101.3kPa+(1000kg/m³)⋅(9.81m/s²)⋅(1.50m)
Solve for P:
P≈101.3kPa+14,715Pa≈116,015Pa≈116.02kPa
So, the absolute pressure at the bottom of the pool is approximately 116.02 kPa.
(b) To find the pressure increase at the bottom of the pool after two persons with a combined mass of 150 kg enter and float, we need to consider the additional weight of the water displaced by the persons. This additional weight will create an increase in pressure at the bottom of the pool.
The buoyant force (Fb) on an object submerged in a fluid is given by Archimedes' principle:
Fb =ρ⋅V⋅g
Where:
Fb is the buoyant force.
ρ is the density of the fluid (water), which is 1000 kg/m³.
V is the volume of water displaced, which is equal to the volume of the submerged part of the persons.
g is the acceleration due to gravity (9.81 m/s²).
The volume of water displaced is equal to the volume of the persons submerged, which is given by the cross-sectional area (A) of the pool base times the depth (h) to which they are submerged:
V=A⋅h
A=π⋅r^2
Given that the radius (r) is 3.00 m and the depth (h) is 1.50 m, we can calculate the area and then the volume:
A=π⋅(3.00m)^2 ≈28.27m²
V=28.27m²⋅1.50m≈42.41m³
Now, we can calculate the buoyant force:
Fb=(1000kg/m³)⋅(42.41m³)⋅(9.81m/s²)≈416,068N
Since pressure is force per unit area, we need to calculate the additional pressure (ΔP) at the bottom of the pool:
ΔP= Fb/A
A=π⋅(3.00m)^2 ≈28.27m²
ΔP= 416,068N / 28.27m² ≈14,699Pa≈0.147kPa
So, the pressure increase at the bottom of the pool after the persons enter and float is approximately 0.147 kPa.
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An object is 12 m long, 0.65 m wide, and 13 cm high. Calculate the volume of this object.
The volume corresponds to the measure of the space occupied by a body. From the given dimensions we can intuit that we are looking to find the Volume of an Cuboid, that is, an orthogonal rectangular prism, whose faces form straight dihedral angles.
Mathematically the volume of this body is given as
[tex]V = lWh[/tex]
Where,
L = Length
W = Width
H = High
[tex]V = (12)(0.65)(13*10^{-2})[/tex]
[tex]V = 1.014m^3[/tex]
Note: The value given for the height was in centimeters, so it was transformed to meters.
A cart of mass 250 g is place on a frictionless horizontal air track. A spring having a spring constant of 9.5 N/m is attached between the cart and the left end o f the track. If the cart is displaced 4.5 cm from its equilibrium position, find: a) The period at which it oscillates.[1.0 s] b) Its maximum speed.[0.28 m/s] c) Its speed when it is located 2.0 cm from its equilibrium position.[0.25 m/s]
Answer:
a) [tex]T=1.02 s[/tex]
b) [tex]v=0.28\frac{m}{s}[/tex]
c) [tex]v=0.25\frac{m}{s}[/tex]
Explanation:
a) The period is defined as:
[tex]T=\frac{2\pi}{\omega}[/tex]
[tex]\omega[/tex] is the natural frequency of the system, in this case is given by:
[tex]\omega=\sqrt{\frac{k}{m}}\\\omega=\sqrt{\frac{9.5\frac{N}{m}}{250*10^{-3}kg}}\\\omega=6.16\frac{rad}{s}[/tex]
Now, we calculate the period:
[tex]T=\frac{2\pi}{6.16\frac{rad}{s}}\\T=1.02 s[/tex]
b) According to the law of conservation of energy, we have:
[tex]\frac{kx^2}{2}=\frac{mv^2}{2}\\v=\sqrt{\frac{kx^2}{m}}\\v=\sqrt{\frac{(9.5\frac{N}{m})(4.5*10^{-2}m)^2}{250*10^{-3}kg}}\\v=0.28\frac{m}{s}[/tex]
c) In this case, we have:
[tex]U_i=U_f+K_f\\\frac{kx_i^2}{2}=\frac{kx_f^2}{2}+\frac{mv^2}{2}\\v=\sqrt{\frac{k(x_i^2-x_f^2)}{m}}\\v=\sqrt{\frac{9.5\frac{N}{m}((4.5*10^{-2}m)^2-(2*10^{-2}m)^2)}{250*10^{-3}kg}}\\v=0.25\frac{m}{s}[/tex]
The maximum speed is; 0.28 m/s
The speed when it is located 2.0 cm from its equilibrium position is; 0.25 m/s
What is the speed from the equilibrium position?A) Formula for period is;
T = √(2π/ω)
where ω which is the natural angular frequency
ω = √(k/m)
we are given;
k = 9.5 N/m
m = 250 g = 0.25 kg
x = 4.5 cm = 0.045 m
Thus;
ω = √(9.5/0.25)
ω = 6.1 rad/s
Thus;
Period is; T = √(2π/6.1)
T = 1.02 s
B) From law of conservation of energy;
maximum speed is;
v = √(kx²/m)
v = √(9.5 * 0.045²/0.25)
v = 0.28 m/s
C) We are now given;
x₁ = 4.5cm = 0.045 m
x₂ = 2 cm = 0.02 m
Thus, speed is gotten from;
v = √((k(x₁ - x₂)/m)
v = √((9.5(0.045 - 0.02)/0.25)
v = 0.25 m/s
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A 2.3 kg block executes SHM while attached to a horizontal spring of spring constant 150 N/m. The maximum speed of the block as it slides on a horizontal frictionless surface is 2.3 m/s. What are (a) the amplitude of the block's motion, (b) the magnitude of its maximum acceleration, and (c) the magnitude of its minimum acceleration? (d) How long does the block take to complete 4.1 cycles of its motion?
Answer
given,
mass of block = m = 2.3 Kg
spring constant = k = 150 N/m
speed = 2.3 m/s
a) we know,
[tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega = \sqrt{\dfrac{150}{2.3}}[/tex]
ω = 8.08 rad/s
v = A ω
2.3 = A x 8.08
A = 0.285 m
b) maximum acceleration
a = A ω²
a = 0.285 x 8.08²
a = 18.58 m/s²
c) acceleration is minimum at mean position the minimum value of acceleration is zero.
d) time period
[tex]T = \dfrac{2\pi}{\omega}[/tex]
[tex]T = \dfrac{2\pi}{8.08}[/tex]
T = 0.778 s
t = 4.1 x 0.7778
t = 3.188 s
You are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the ground. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 270N .
Part A
How much work do you do on the refrigerator during its trip down the ramp?
W = ???
Final answer:
The work done on the refrigerator during its trip down the ramp is 1350 J.
Explanation:
To calculate the work done while unloading the refrigerator, we need to determine the displacement of the refrigerator and the force applied.
The displacement is given by the length of the ramp, which is 5.0 m.
The force applied is given as 270 N.
To calculate the work done, we use the formula:
Work = Force x Displacement x Cosine(angle)
In this case, the angle is 0 degrees since the force is applied horizontally.
Therefore, the work done on the refrigerator is:
Work = 270 N x 5.0 m x Cos(0°) = 1350 J.
An electromagnetic wave is transporting energy in the negative y direction. At one point and one instant the magnetic field is in the positive x direction. The electric field at that point and instant is:Why is (D) the answer? I thought it's (C)A) positive y directionB) negative y directionC) positive z directionD) negative z direction
Answer:negative Z direction
Explanation:
It is given that EM wave is travelling in negative y direction i.e. -[tex]\hat{j}[/tex]
Magnetic Field is in positive x direction i.e. [tex]\hat{i}[/tex]
We know that Electric field and magnetic field are perpendicular to each other and they are mutually perpendicular to direction of wave propagation.
Mathematically if Electric field is in negative z direction it will yield the direction of wave propagation
[tex]=(-\hat{k})\times \hat{i}[/tex]
[tex]=-\hat{j}[/tex]