Answer:
Part a)
[tex]W = 1.2 \times 10^5 J[/tex]
Part b)
[tex]Q = 5.6 \times 10^5 J[/tex]
Explanation:
It given that player will feel exhausted when he is using his internal energy of [tex]8.0 \times 10^5 J[/tex]
PART a)
it is given that heat loss by the player is given as
[tex]Q = 6.8 \times 10^5 J[/tex]
now by first law of thermodynamics we have
[tex]\Delta U = Q + W[/tex]
now we have
[tex]8.0 \times 10^5 = 6.8 \times 10^5 + W[/tex]
[tex]W = 1.2 \times 10^5 J[/tex]
PART b)
It is given that another player did the work as
[tex]W = 2.4 \times 10^5 J[/tex]
now we have first law of thermodynamics
[tex]\Delta U = Q + W[/tex]
now we have
[tex]8.0 \times 10^5 = 2.4 \times 10^5 + Q[/tex]
[tex]Q = 5.6 \times 10^5 J[/tex]
Using the first law of thermodynamics, we find that the first player has done -1.2 x 10^5 J of work and the second player has lost 10.4 x 10^5 J of heat.
Explanation:The questions posed are about applying the concept of the first law of thermodynamics in determining the amount of work done by a football player and the heat lost in the process. This law is also known as the law of energy conservation, and it can be written as ΔU = Q - W, where ΔU is the change in the system's internal energy, Q is the heat added to the system, and W is the work done by the system.
(a) The player who was dressed too lightly had to leave the game after losing 6.8 × 105 J of heat. Using the first law of thermodynamics, we can calculate the work done by the player. If the change in the internal energy before the player gets exhausted is 8.0 × 105 J (the energy used) and the heat lost is 6.8 × 105 J, then the work done (W) can be calculated as follows:
W = Q - ΔU = 6.8 × 105 J - 8.0 × 105 J = -1.2 × 105 J
(b) Another player was able to do 2.4 × 105 J of work before getting exhausted. The magnitude of the heat that he lost can then be calculated as follows:
Q = ΔU + W = 8.0 × 105 J + 2.4 × 105 J = 10.4 × 105 J
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The specific heat of copper is 385 J/kg·K. If a 2.6 kg block of copper is heated from 340 K to 450 K, how much thermal energy is absorbed?
Answer:
Heat absorbed, Q = 110110 J
Explanation:
It is given that,
The specific heat of copper is, [tex]c=385\ J/kg.K[/tex]
Mass of block, m = 2.6 kg
Initial temperature, [tex]T_1=340\ K[/tex]
Final temperature, [tex]T_2=450\ K[/tex]
Thermal energy is given by :
[tex]Q=mc\Delta T[/tex]
[tex]Q=mc(T_2-T_1)[/tex]
[tex]Q=2.6\times 385\times (450-340)[/tex]
Q = 110110 J
So, the thermal heat of 110110 J is absorbed. Hence, this is the required solution.
bullet of mass 0.02070 �� collides inelastically with a wooden block of mass 28.00 ��, initially at rest. After the collision the, system has speed of 1.180 �/�. (a) What was initial speed of the bullet
Explanation:
It is given that,
Mass of bullet, m₁ = 0.0207 kg
Mass of wooden block, m₂ = 28 kg
Wooden block is initially at rest, u₂ = 0 m/s
After the collision the, system has speed of, v = 1.180 m/s
We need to find the initial speed (u₁) of the bullet. It can be calculated using the conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
[tex]0.0207\ kg\times u_1+0=(0.0207\ kg+28\ kg)\times 1.180\ m/s[/tex]
[tex]0.0207\times u_1=33.06[/tex]
[tex]u_1=1597.10\ m/s[/tex]
So, the initial speed of the bullet is 1597.10 m/s. Hence, this is the required solution.
A chemistry student needs of 40mL diethylamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of diethylamine is 0.7 . Calculate the mass of diethylamine the student should weigh out. Round your answer to significant digits.
Explanation:
Mass = density × volume
m = (0.7 g/mL) × (40 mL)
m = 28 g
Rounding to 1 significant figure, the student should weigh out 30 grams of diethylamine.
By using the formula Mass = Density x Volume, we find that the student needs to weigh out 28 grams of diethylamine for 40mL volume, as the density of diethylamine is 0.7g/mL.
Explanation:The question asks for the mass of diethylamine the student should weigh out to obtain 40mL of diethylamine. Diethylamine has a density of 0.7 g/mL, according to the CRC Handbook of Chemistry and Physics. The formula to find mass when you have volume and density is:Mass = Density x Volume. Therefore, to find the mass of the diethylamine, we multiply the volume (40 mL) by the density (0.7 g/mL), which equals 28 grams. So, the student should weigh out 28 grams of diethylamine.
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A simple model for a person running the 100 m dash is to assume the sprinter runs with constant acceleration until reaching top speed, then maintains that speed through the finish line. If a sprinter reaches his top speed of 11.2 m/s in 2.14 s, what will be his total time?
Final answer:
The sprinter's total time in the 100 m dash is 10 seconds.
Explanation:
In this problem, the sprinter reaches his top speed of 11.2 m/s in 2.14 seconds.
To find the total time, we need to consider two parts: the time it takes to reach the top speed and the time it takes to cover the remaining distance at that speed.
First, calculate the time it takes to reach the top speed using the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values: 11.2 = 0 + a(2.14), we can solve for a to find a = 5.23 m/s².
Now, to find the time it takes to cover the remaining distance at the top speed, we can use the equation:
t = d / v
where d is the distance and v is the velocity.
Since the total distance is 100 m, subtracting the distance covered during the acceleration phase, we get the remaining distance as 100 - (0.5)(5.23)(2.14)² = 88.2 m.
Dividing this distance by the top speed of 11.2 m/s, we find t = 88.2 / 11.2 = 7.86 s.
To find the total time, we add the time it takes to reach the top speed and the time it takes to cover the remaining distance, so the sprinter's total time is 2.14 s + 7.86 s = 10 s.
A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 21.6 m/s. (a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.) kg · m/s (b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist. N
(a) +9.30 kg m/s
The impulse exerted on an object is equal to its change in momentum:
[tex]I= \Delta p = m \Delta v = m (v-u)[/tex]
where
m is the mass of the object
[tex]\Delta v[/tex] is the change in velocity of the object, with
v = final velocity
u = initial velocity
For the volleyball in this problem:
m = 0.272 kg
u = -12.6 m/s
v = +21.6 m/s
So the impulse is
[tex]I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s[/tex]
(b) 155 N
The impulse can also be rewritten as
[tex]I=F \Delta t[/tex]
where
F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)
[tex]\Delta t[/tex] is the duration of the collision
In this situation, we have
[tex]\Delta t = 0.06 s[/tex]
So we can re-arrange the equation to find the magnitude of the average force:
[tex]F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N[/tex]
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 81.1 V/m. Calculate the intensity S of this wave. Then, determine the amount of energy U that flows through the area of 0.0253 m^2 over an interval of 19.9 s, assuming that the area is perpendicular to the direction of wave propagation.
Answer:
U = 4.39 J
Explanation:
Electric field energy stored in the medium or vacuum is given as
[tex]U = \frac{1}{2}\epsilon_0 E^2 V[/tex]
here we know that
[tex]\epsilon_0 = 8.85 \times 10^{-12} [/tex]
E = 81.1 V/m
V = volume
[tex]V = (0.0253)(speed \times time)[/tex]
[tex]V = (0.0253)(3\times 10^8 \times 19.9)[/tex]
[tex]V = 1.51 \times 10^8 m^3[/tex]
now from above formula we have
[tex]U = \frac{1}{2}(8.85 \times 10^{-12})(81.1)^2(1.51 \times 10^8)[/tex]
[tex]U = 4.39 J[/tex]
The nuclear potential that binds protons and neutrons in the nucleus of an atom is often approximated by a square well. Imagine a proton conned in an innite square well of length 105 nm, a typical nuclear diameter. Calculate the wavelength and energy associated with the photon that is emitted when the proton undergoes a transition from the rst excited state (n 2) to the ground state (n 1). In what region of the electromagnetic spectrum does this wavelength belong?
3. The nuclear potential that binds protons and neutrons in the nucleus of an atom
is often approximated by a square well. Imagine a proton confined in an infinite
square well of length 10−5 nm, a typical nuclear diameter. Calculate the wavelength
and energy associated with the photon that is emitted when the proton undergoes a
transition from the first excited state (n = 2) to the ground state (n = 1). In what
region of the electromagnetic spectrum does this wavelength belong?
Answer 3
We are given that,
Length of square well = L = 10−5
nm = 10−14 m.
Energy of proton in state n is given by,
En =
π
2n
2~
2
2mpL2
,
where L is the width of the square well.
⇒ E1 =
π
2~
2
2mpL2
E2 =
4π
2~
2
2mpL2
·
A 11 kg mass on a frictionless inclined surface is connected to a 2.1 kg mass. The pulley is massless and frictionless, and the connecting string is massless and does not stretch. The 2.1 kg mass is acted upon by an upward force of 6.6 N, and thus has a downward acceleration of only 3.6 m/s 2 . The acceleration of gravity is 9.8 m/s 2 . What is the tension in the connecting string?
Answer:
6.4 N
Explanation:
Draw free body diagrams for each mass.
For the mass on the inclined surface, the sum of forces parallel to the incline are:
∑F = ma
T - Mg sin θ = Ma
For the hanging mass, the sum of the forces in the y direction are:
∑F = ma
T - mg + F = m(-a)
We're given that g = 9.8 m/s², M = 11 kg, m = 2.1 kg, F = 6.6 N, and a = 3.6 m/s².
From the second equation, we have everything we need to find the tension force.
T - (2.1)(9.8) + 6.6 = (2.1)(-3.6)
T = 6.42 N
Rounded to 2 sig figs, the tension is 6.4 N.
A ball is shot straight up from the surface of the earth with an initial speed of 19.6 m/s. Neglect any effects due to air resistance. How much time elapses between the throwing of the ball and its return to the original launch point?
Answer:
4 s
Explanation:
u = 19.6 m/s, g = 9.8 m /s^2
Let the time taken to reach the maximum height is t.
Use first equation of motion.
v = u + at
At maximum height, final velocity v is zero.
0 = 19.6 - 9.8 x t
t = 19.6 / 9.8 = 2 s
As the air resistance be negligible, is time taken to reach the ground is also 2 sec.
So, total time taken be the ball to reach at original point = 2 + 2 = 4 s
The ball takes 4 seconds to return to its original launch point.
Explanation:To find the time it takes for the ball to return to its original launch point, we need to determine the total time it takes for the ball to reach its maximum height and come back down. The ball is shot straight up, which means its initial velocity is positive and its final velocity is negative (when it returns to the launch point). We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed. In this case, the final velocity is -19.6 m/s (negative because it's going down), the initial velocity is 19.6 m/s, and the acceleration is -9.8 m/s^2 (due to gravity). Plugging the values into the equation and solving for t:
v = u + at
-19.6 = 19.6 - 9.8t
-39.2 = -9.8t
t = -39.2 / -9.8
t = 4 seconds
It takes 4 seconds for the ball to return to its original launch point.
A car moves a distance of 50.0 km West, followed by a distance of 64.9 km North What was the magnitude of the displacement of the car, in units of kilometers?
Answer:
Displacement of the car, XY = 81.92 km
Explanation:
From the attached figure,
OX = 64.9 km (in north direction)
OY = 50 km (in west direction)
We have to find the displacement of the car. The shortest path covered by a car is called displacement of the car. Here, XY shows the displacement of the car :
Using Pythagoras equation as :
[tex]XY^2=OX^2+OY^2[/tex]
[tex]XY^2=(64.9\ km)^2+(50\ km)^2[/tex]
XY = 81.92 km
Hence, the displacement of the car is 81.92 km.
The magnitude of the car's displacement is calculated using the Pythagorean theorem by treating the westward and northward movements as the sides of a right-angled triangle. The displacement, representing the hypotenuse of that triangle, is found to be approximately 81.93 km.
The student has asked to find the magnitude of displacement of a car that moves 50.0 km West and then 64.9 km North.
Calculating Displacement
To calculate the resultant displacement, we treat the distances as vector quantities and use the Pythagorean theorem. The car's westward and northward movements are at right angles to each other, so we can draw this as a right-angled triangle where the westward distance is one side (50.0 km), the northward distance is the other side (64.9 km), and the hypotenuse will be the displacement.
Using the Pythagorean theorem:
Displacement = \/(westward distance)^2 + (northward distance)^2Displacement = \/(50.0 km)^2 + (64.9 km)^2Displacement = \/(2500 + 4212.01) km^2Displacement = \/(6712.01) km^2Displacement = 81.93 kmThe magnitude of the car's displacement, therefore, is approximately 81.93 km.
Describe Lenz's law.
Answer:
Explanation:
Lenz law is used to find the direction of induced emf in the coil.
It state taht the direction of induced emf in the coil is such that it always opposes the change due to which it is produced.
Suppose there is a coil and a north pole of the magnet comes nearer to the coil. Due to changing magnetic flux an induced emf is developed in the coil whose direction is such that the north pole moves away. That means this face of the coil behaves like a north pole and the current flows at this face is in anticlockwise wise direction.
Final answer:
Lenz's law is a manifestation of the conservation of energy in physics. It states that the direction of the induced electromotive force (emf) drives current around a wire loop to always oppose the change in magnetic flux that causes the emf. Lenz's law ensures that the induced current produces a magnetic field that tries to cancel out the change in flux caused by a changing magnetic field.
Explanation:
Lenz's law is a manifestation of the conservation of energy in physics. It states that the direction of the induced electromotive force (emf) drives current around a wire loop to always oppose the change in magnetic flux that causes the emf. This means that when there is a change in the magnetic field through a circuit, the induced current will create a magnetic field that acts against the change.
For example, if a magnet is brought near a wire loop, the induced current will flow in such a way that it creates a magnetic field that opposes the motion of the magnet. This is because the changing magnetic field induces an emf in the wire loop, and Lenz's law ensures that the induced current produces a magnetic field that tries to cancel out the change in flux caused by the magnet.
You connect a voltmeter to an unknown resistor in a circuit in order to measure the potential difference across that resistor. How should you connect it to the resistor and what would the voltmeter's ideal internal resistance be? A. In series Fint B. In parallel; int- O C. In series; rint- 0 D. In parallel; rint →
Answer:
option (a)
Explanation:
To make a galvanometer into voltmeter, we have to connect a high resistance in series combination.
The voltmeter is connected in parallel combination with teh resistor to find the voltage drop across it.
An ideal voltmeter has very high resistance that means it has a resistance as infinity.
Answer:
A. In series Fint
Explanation:
If you connect it to the resistor, the voltmeter's ideal internal resistance be In series Fint.
The temperature in degrees Fahrenheit in terms of the Celsius temperature is given by . The temperature in degrees Celsius in terms of the Kelvin temperature is given by . Write a formula for the temperature in degrees Fahrenheit in terms of the Kelvin temperature . It is not necessary to simplify.
Answer:
The temperature in degrees Fahrenheit in terms of the Celsius temperature is given by .
As we know by the linear relation
[tex]\frac{^o F - 32}{212 - 32} = \frac{^o C - 0}{100 - 0}[/tex]
now we have
[tex]^o F - 32 = \frac{180}{100} ^o C[/tex]
so we have
[tex]^o F = \frac{9}{5} ^o C + 32[/tex]
The temperature in degrees Celsius in terms of the Kelvin temperature is given by
As we know by the linear relation
[tex]\frac{^o C - 0}{100 - 0} = \frac{K - 273}{373 - 273}[/tex]
now we have
[tex]C = K - 273[/tex]
the temperature in degrees Fahrenheit in terms of the Kelvin temperature .
As we know by the linear relation
[tex]\frac{^o F - 32}{212 - 32} = \frac{K - 273}{373 - 273}[/tex]
now we have
[tex]^o F - 32 = \frac{180}{100} (K - 273)[/tex]
so we have
[tex]^o F = \frac{9}{5} (K - 273) + 32[/tex]
A tension force of 165 N inclined at 35.0° above the horizontal is used to pull a 31.0 kg storage crate a distance of 5.60 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b) the coefficient of kinetic friction between the crate and surface. HINT (a) the work done by the tension force (in J) J (b) the coefficient of kinetic friction between the crate and surface
(a) 756.9 J
The work done by a force when moving an object is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement of the object
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this situation,
F = 165 N
d = 5.60 m
[tex]\theta=35.0^{\circ}[/tex]
so the work done by the tension is
[tex]W=(165 N)(5.60 m)cos 35.0^{\circ}=756.9 J[/tex]
(b) 0.646
The crate is moving at constant speed: this means that the acceleration of the crate is zero, so the net force on the crate is also zero.
There are only two forces acting on the crate along the horizontal direction:
- The horizontal component of the tension, [tex]Tcos \theta[/tex], forward
- The frictional force, [tex]-\mu N[/tex], backward, with [tex]\mu[/tex] being the coefficient of kinetic friction, N being the normal reaction of the surface on the crate
The normal reaction is the sum of the weight of the crate + the vertical component of the tension, so
[tex]N=mg-T sin \theta = (31.0 kg)(9.8 m/s^2) - (165 N)sin 35.0^{\circ}=209.2 N[/tex]
Since the net force is zero, we have
[tex]T cos \theta-\mu N =0[/tex]
where
T = 165.0 N
[tex]\theta=35.0^{\circ}[/tex]
N = 209.2 N
Solving for [tex]\mu[/tex], we find the coefficient of friction:
[tex]\mu = \frac{T cos \theta}{N}=\frac{(165.0 N)(cos 35^{\circ})}{209.2 N}=0.646[/tex]
Final answer:
In this physics problem, we determine the work done by a tension force inclining at an angle above the horizontal and find the coefficient of kinetic friction when a crate moves at a constant speed.
Explanation:
A tension force of 165 N inclined at 35.0° above the horizontal is used to pull a 31.0 kg storage crate at a distance of 5.60 m on a rough surface.
(a) Work done by the tension force: To calculate work, use the formula W = Fd cos(θ), where F is the force, d is the distance moved, and θ is the angle between the force and displacement. Here, W = 165 N * 5.60 m * cos(35.0°).
(b) Coefficient of kinetic friction: Since the crate moves at a constant speed, the work done by the tension force equals the work done against friction. Find the frictional force using the work done formula and then the coefficient of kinetic friction.
A rock is thrown vertically upward with a speed of 18.0 m/s from the roof of a building that is 50.0 m above the ground. Assume free fall : Part A) In how many seconds after being thrown does the rock strike the ground? Part B) What is the speed of the rock just before it strikes the ground?
The rock strikes the ground after approximately 3.67 seconds. The speed of the rock just before it strikes the ground is approximately 17.0 m/s.
Explanation:To find the time it takes for the rock to strike the ground, we can use the equation for vertical motion. Assuming negligible air resistance, the initial velocity is 18.0 m/s and the vertical displacement is 50.0 m.
Using the equation y = yo + vyo*t - 1/2*g*t^2, where y is the final position, yo is the initial position, vyo is the initial velocity, g is the acceleration due to gravity, and t is the time, we can solve for t. Plugging in the values, we get:
50.0 = 0 + 18.0*t - 1/2*9.8*t^2
After rearranging the equation and solving the quadratic equation, we find that t = 3.67 seconds.
To find the speed of the rock just before it strikes the ground, we can use the equation for vertical motion. The final velocity v is equal to the initial velocity vyo - g*t. Plugging in the values, we get:
v = 18.0 - 9.8*3.67
Calculating the value, we find that the speed of the rock just before it strikes the ground is approximately 17.0 m/s.
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A ball is thrown straight up with an initial speed of 16.9 m/s. At what height above its initial position will the ball have one‑half its initial speed?
Answer:
10.9 m
Explanation:
We can solve the problem by using the law of conservation of energy.
The initial mechanical energy is just the kinetic energy of the ball:
[tex]E = K_i = \frac{1}{2}mu^2[/tex]
where m is the mass of the ball and u = 16.9 m/s the initial speed.
At a height of h, the total mechanical energy is sum of kinetic energy and gravitational potential energy:
[tex]E=K_f + U_f = \frac{1}{2}mv^2 + mgh[/tex]
where v is the new speed, g is the gravitational acceleration, h is the height of the ball.
Due to the conservation of energy,
[tex]\frac{1}{2}mu^2 = \frac{1}{2}mv^2 +mgh\\u^2 = v^2 + 2gh[/tex] (1)
Here, at a height of h we want the speed to be 1/2 of the initial speed, so
[tex]v=\frac{1}{2}u[/tex]
So (1) becomes
[tex]u^2 = (\frac{u}{2})^2+2gh\\\frac{3}{4}u^2 = 2gh[/tex]
So we can find h:
[tex]h=\frac{3u^2}{8g}=\frac{3(16.9 m/s)^2}{8(9.8 m/s^2)}=10.9 m[/tex]
Final answer:
To find the height where the ball has one-half its initial speed, we can use the equations vf = v0 + gt and d = v0t - 0.5gt2.
Explanation:
To find the height above its initial position where the ball has one-half its initial speed, we need to use the fact that the initial velocity (v0) of the ball is 16.9 m/s. At the highest point of the ball's trajectory, the velocity will be zero. We can use the formula vf = v0 + gt, where vf is the final velocity, g is the acceleration due to gravity, and t is the time it takes for the ball to reach its highest point.
By substituting vf = 0 and v0 = 16.9 m/s, we can solve for t. Once we have the value of t, we can use the equation d = v0t - 0.5gt2 to calculate the height (d) above the initial position where the ball will have one-half its initial speed.
By substituting the calculated value of t into the equation, we can find the value of d.
he electric fux through a surface is zero. Thereloee there are no chargm inside the ace A. True, if there in no fus theee ran be no chargs B. Fase, flux hae nothing to do with nclosed charge C. True, thst is how Ewoks control the world banks D. Fabe, the sum of all thargrs inside can be aro
Answer:
Option (D)
Explanation:
According to the Gauss theorem in electrostatics, the electric flux passing through any surface is equal to the one divided by epsilon note ties the total cahrge enclosed in the surface.
As the flux is zero it means the enclosed charge is zero. It means the sum of the total cahrge inside is zero.
When during new product development is Design For Manufacture and Assembly (DFMA) most effective? a) At all times b) During production c) During process design and development d) During product design and development e) Before design
Answer:most likely E
Explanation:
Why would anybody do something after design there done with there work after that
A 51.0 kg cheetah accelerates from rest to its top speed of 31.7 m/s. HINT (a) How much net work (in J) is required for the cheetah to reach its top speed? J (b) One food Calorie equals 4186 J. How many Calories of net work are required for the cheetah to reach its top speed? Note: Due to inefficiencies in converting chemical energy to mechanical energy, the amount calculated here is only a fraction of the energy that must be produced by the cheetah's body. Cal
(a) [tex]2.56\cdot 10^4 J[/tex]
The work-energy theorem states that the work done on the cheetah is equal to its change in kinetic energy:
[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]
where
m = 51.0 kg is the mass of the cheetah
u = 0 is the initial speed of the cheetah (zero because it starts from rest)
u = 31.7 m/s is the final speed
Substituting, we find
[tex]W=\frac{1}{2}(51.0 kg)(31.7 m/s)^2 - \frac{1}{2}(51.0 kg)(0)^2=2.56\cdot 10^4 J[/tex]
(b) 6.1 cal
The conversion between calories and Joules is
1 cal = 4186 J
Here the energy the cheetah needs is
[tex]E=2.56\cdot 10^4 J[/tex]
Therefore we can set up a simple proportion
[tex]1 cal : 4186 J = x : 2.56\cdot 10^4 J[/tex]
to find the equivalent energy in calories:
[tex]x=\frac{(1 cal)(2.56\cdot 10^4 J)}{4186 J}=6.1 cal[/tex]
The cheetah needs 25,972.35 Joules or 6.2 Calories of work to reach its top speed of 31.7 m/s.
Explanation:The cheetah accelerates from a state of rest to its top speed. This involves work done on the cheetah which we can compute using the formula for kinetic energy, as work done is equal to change in kinetic energy. The initial speed of the cheetah is 0, so initial kinetic energy is also 0. The final kinetic energy when the cheetah is at its top speed is ½ m v^2, where m is the mass and v is the speed.
(a) So, the work done, W = ½ * 51.0kg * (31.7 m/s)^2 = 25972.35 J,
(b) To convert Joules into Calories, we use 1 Calorie = 4186 J, so the Calories of work done = 25972.35 J / 4186 = 6.2 Calories. Note that due to inefficiencies in the energy conversion process in bodies, the actual energy produced by the cheetah's body will be more than 6.2 Calories to reach this speed.
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A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field E⃗ = (100 N/C) iˆ . Calculate the flux of the net electric field through a Gaussian sphere of radius R = 10.0 cm centered at the origin. (ε 0 = 8.85 × 10-12 C2/N · m2)
Answer:
The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]
Explanation:
Given that,
Charge [tex]q= 2.00 \mu C[/tex]
Electric field E = 100 N/C i
Radius R = 10.0 cm
We need to find the flux. It can be calculate using Gauss's law
The flux of the net electric field
[tex]\phi=\dfrac{q}{\epsilon_{0}}[/tex]
[tex]\phi=\dfrac{2.00\times10^{-6}}{8.85\times10^{-12}}[/tex]
[tex]\phi=2.26\times10^{5}\ Nm^2/C[/tex]
Hence, The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]
a shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. the shot hits the ground 2.08 s later. you can ignore air resistance. how far did she throw the shot?
Answer:
15.7 m
Explanation:
The range (horizontal distance) of the projectile is determined only by its horizontal motion.
The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:
[tex]v_x = v cos \theta[/tex]
where
v = 12.0 m/s is the initial velocity
[tex]\theta=51.0^{\circ}[/tex] is the angle between the direction of v and the horizontal
Substituting,
[tex]v_x = (12.0 m/s)(cos 51.0^{\circ} )=7.55 m/s[/tex]
We know that the projectile hits the ground in a time of
t = 2.08 s
so the horizontal distance covered is
[tex]d = v_x t = (7.55 m/s)(2.08 s)=15.7 m[/tex]
Final answer:
The shot putter threw the shot approximately 15.71 meters by using the horizontal component of the initial velocity and the time of flight.
Explanation:
To calculate the distance the shot putter threw the shot, we need to break down the velocity into its horizontal and vertical components. Given that the shot was released with a velocity of 12.0 m/s at an angle of 51.0 degrees above the horizontal, we can use trigonometry to find these components.
The horizontal velocity (vx) is given by vx = v * cos(θ) = 12.0 m/s * cos(51.0) = 7.55 m/s. The vertical velocity (vy) is vy = v * sin(θ) = 12.0 m/s * sin(51.0) = 9.35 m/s.
Since there's no acceleration in the horizontal direction (ignoring air resistance), this component of the velocity will remain constant until the shot hits the ground. Thus, the horizontal distance (range) the shot travels is simply the product of the horizontal velocity and the time it's in the air, given by Range = vx * t = 7.55 m/s * 2.08 s = 15.71 meters. So, the shot putter threw the shot approximately 15.71 meters.
Assume your car reaches a speed of 21.7 m/s at a steady rate for 5.05 s after the light turns green. (a) What distance have you traveled during this time? (b) What is your average speed during this time?
Answer:
The distance and average speed are 54.79 m and 10.85 m.
Explanation:
Given that,
Speed = 21.7 m/s
Time = 5.05 s
(a). We need to calculate the distance
Firstly we will find the acceleration
Using equation of motion
[tex]v = u+at[/tex]
[tex]a = \dfrac{v-u}{t}[/tex]
Where, v = final velocity
u = initial velocity
t = time
Put the value in the equation
[tex]a = \dfrac{21.7-0}{5.05}[/tex]
[tex]a = 4.297 m/s^2[/tex]
Now, using equation of motion again
For distance,
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
[tex]s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2[/tex]
[tex]s=54.79\ m[/tex]
The distance is 54.79 m.
(b). We need to calculate the average speed during this time
[tex]v_{avg}=\dfrac{D}{T}[/tex]
Where, D = total distance
T = time
Put the value into the formula
[tex]v_{avg}=\dfrac{54.79}{5.05}[/tex]
[tex]v_{avg}=10.85\ m/s[/tex]
Hence, The distance and average speed are 54.79 m and 10.85 m.
A woman walks 237 m in the direction 27.2 east of north, then 335 m directly east. Find the difference between the distance the woman walks and the magnitude of her displacement. (Answer in m.)
Answer:
81.1 m
Explanation:
X = 237 m 27.2 east of north 237 (Sin27.2 i + Cos 27.2 j)
X = 108.33 i + 210.8 j
Y = 335 east = 335 i
Displacement = X + Y = 108.33 i + 210.8 j + 335 i = 443.33 i + 210.8 j
(magnitude of displacement)^2 = 443.33^2 + 210.8^2
magnitude of displacement = 490.9 m
Distance traveled = 237 + 335 = 572 m
Difference in the magnitude of displacement and distance = 572 - 490.9
= 81.1 m
A person takes a trip, driving with a constant speed of 92.5 km/h, except for a 28.0-min rest stop. The person's average speed is 72.2 km/h. How much time is spent on the trip?
Answer:
40.3 min
Explanation:
First of all, let's convert every quantity into SI units:
[tex]v_1 = 92.5 km/h = 25.7 m/s[/tex] (speed in the first part of the trip)
[tex]t_2 = 28.0 min = 1680 s[/tex] time during which the person has stopped
[tex]v=72.2 km/h = 20.1 m/s[/tex] (average speed of the whole trip)
The average speed is the ratio between the total distance covered, d, and the total time taken, t:
[tex]v=\frac{d}{t}[/tex] (1)
The total distance covered is simply
[tex]d = v_1 t_1[/tex]
where [tex]t_1[/tex] is the time during which the person has moved at 92.5 km/h.
The total time taken is
[tex]t= t_1 + t_2[/tex]
So (1) becomes
[tex]v=\frac{v_1 t_1}{t_1 + t_2}[/tex]
Solving for [tex]t_1[/tex]:
[tex]v t_1 + v t_2 = v_1 t_1\\vt_2 = (v_1+v)t_1\\t_1 = \frac{v t_2}{v_1+v}=\frac{(20.1 m/s)(1680 s)}{25.7 m/s + 20.1 m/s}=737.3 s[/tex]
which corresponds to
[tex]t_2 = 737.3 s = 12.3 min[/tex]
So the total time of the trip is
[tex]t = 28.0 min + 12.3 min = 40.3 min[/tex]
A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second (1) 33 % . is (2) 40 % (4) 77 % / (3) 66 % A particle is thrown upward from groun
Answer:
Option 2 is the correct answer.
Explanation:
We have equation of motion s = ut + 0.5at²
Here u = 0 m/s
So, s = 0.5at²
Distance traveled in first second = 0.5 x a x 1² = 0.5 a
Distance traveled in second second = 0.5 x a x 2² - 0.5 x a x 1²= 1.5 a
Distance traveled in third second = 0.5 x a x 3² - 0.5 x a x 2²= 2.5 a
Distance traveled in fourth second = 0.5 x a x 4² - 0.5 x a x 3²= 3.5 a
The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second
[tex]=\frac{3.5a-2.5a}{2.5a}=0.4=40\%[/tex]
Option 2 is the correct answer.
A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. What is the magnitude of the ball's velocity just before it hits the ground?
The ball's position vector has components
[tex]x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ t[/tex]
[tex]y=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2[/tex]
where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. The ball hits the ground when [tex]y=0[/tex]:
[tex]0=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2\implies t=1.22\,\mathrm s[/tex]
The ball's velocity vector has components
[tex]v_x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ[/tex]
[tex]v_y=\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ-gt[/tex]
so that after 1.22 s, the velocity vector is
[tex]\vec v=(6.13\,\vec\imath-6.79\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]
and the magnitude is
[tex]\|\vec v\|=\sqrt{6.13^2+(-6.79)^2}\,\dfrac{\rm m}{\rm s}=\boxed{9.14\dfrac{\rm m}{\rm s}}[/tex]
A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field?
Answer:
The angle the wire make with respect to the magnetic field is 30°
Explanation:
It is given that,
Length of straight wire, L = 0.6 m
Current carrying by the wire, I = 2 A
Magnetic field, B = 0.3 T
Force experienced by the wire, F = 0.18 N
Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :
[tex]F=ILB\ sin\theta[/tex]
[tex]\theta=sin^{-1}(\dfrac{F}{ILB})[/tex]
[tex]\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\times 0.3\ T})[/tex]
[tex]\theta=30^{\circ}[/tex]
So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.
A satellite orbits Earth at an altitude of 350 km. What is its orbital period? A) 1.1 min B) It can not be determined without the satellite's mass. C) 65 min D) 91 min E) 1440 min
The orbital period of a satellite at an altitude of 350 km above Earth's surface is approximately 1.93 hours or around 115.8 minutes. Given the options, the closest one is option D, with a period of 91 minutes.
Explanation:To answer the question about the satellite orbiting Earth at an altitude of 350 km, we can utilize some physics concepts related to orbital mechanics. Note that a significant factor in determining the orbital period of a satellite (time required for a satellite to complete one orbit) is not the satellite's mass but rather its altitude or, more accurately, the total distance from the center of the Earth, considering the Earth's radius plus the satellite's altitude.
In general, the closer a satellite is to Earth, the shorter its orbital period. The information given indicates that a satellite at an altitude of 350 km will likely have an orbital period around 1.93 hours, which is roughly equivalent to 115.8 minutes. Given that, none of the available options exactly match this result. But the value closest to it would be 91 minutes, which is represented by option D.
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A catapult is tested by Roman legionnaires. They tabulate the results in a papyrus and 2000 years later the archaeological team reads (distances translated into modern units): Range = 0.4 km; angle of launch = π/5 rad; landing height = launch height. What is the initial velocity of launch of the boulders if air resistance is negligible?
Answer:
64.2 m/s
Explanation:
In the x direction:
x = x₀ + v₀ₓ t + ½ at²
400 m = 0 m + v₀ cos (π/5) t + ½ (0 m/s²) t²
t = 400 / (v₀ cos (π/5))
In the y direction:
y = y₀ + v₀ᵧ t + ½ gt²
0 m = 0 m + v₀ sin (π/5) t + ½ (-9.8 m/s²) t²
0 = v₀ sin (π/5) - 4.9 t
t = v₀ sin (π/5) / 4.9
Therefore:
400 / (v₀ cos (π/5)) = v₀ sin (π/5) / 4.9
1960 = v₀² sin (π/5) cos(π/5)
1960 = ½ v₀² sin(2π/5)
3920 / sin(2π/5) = v₀²
v₀ = 64.2 m/s
Answer:
Initial velocity = 423.08m/s
Explanation:
Using formular for Range of projectile
R =V^2 sin2theta/g
Given:
Range=0.4km= 400m
Theta=3.142/5
g= 9.8m/s^2
400= V^2×sin(2×3.142/5)/9.8
400×9.8= V^2Sin 1.26
3920= 0.0219V^2
V^2= 3920/0.0219
V^2= 178995.43
V=sqrt 178995.43
V= 423.08m/s
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).
(a) -2451 N
We can start by calculating the acceleration of the car. We have:
[tex]u=85.0 km/h = 23.6 m/s[/tex] is the initial velocity
v = 0 is the final velocity of the car
d = 125 m is the stopping distance
So we can use the following equation
[tex]v^2 - u^2 = 2ad[/tex]
To find the acceleration of the car, a:
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2[/tex]
Now we can use Newton's second Law:
F = ma
where m = 1100 kg to find the force exerted on the car in order to stop it; we find:
[tex]F=(1100 kg)(-2.23 m/s^2)=-2451 N[/tex]
and the negative sign means the force is in the opposite direction to the motion of the car.
(b) [tex]-1.53\cdot 10^5 N[/tex]
We can use again the equation
[tex]v^2 - u^2 = 2ad[/tex]
To find the acceleration of the car. This time we have
[tex]u=85.0 km/h = 23.6 m/s[/tex] is the initial velocity
v = 0 is the final velocity of the car
d = 2.0 m is the stopping distance
Substituting and solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2[/tex]
So now we can find the force exerted on the car by using again Newton's second law:
[tex]F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N[/tex]
As we can see, the force is much stronger than the force exerted in part a).