You have a glass ball with a radius of 2.00 mm and a density of 2500 kg/m3. You hold the ball so it is fully submerged, just below the surface, in a tall cylinder full of glycerin, and then release the ball from rest. Take the viscosity of glycerin to be 1.5 Pa s and the density of glycerin to be 1250 kg/m3. Use g = 10 N/kg = 10 m/s2. Also, note that the drag force on a ball moving through a fluid is: Fdrag = 6πηrv . (a) Note that initially the ball is at rest. Sketch (to scale) the free-body diagram of the ball just after it is released, while its velocity is negligible. (b) Calculate the magnitude of the ball’s initial acceleration. (c) Eventually, the ball reaches a terminal (constant) velocity. Sketch (to scale) the free-body diagram of the ball when it is moving at its terminal velocity. (d) Calculate the magnitude of the terminal velocity. (e) What is the magnitude of the ball’s acceleration, when the ball reaches terminal velocity? (f) Let’s say that the force of gravity acting on the ball is 4F, directed down. We can then express all the forces in terms of F.

Answer:

5 m/s

Explanation:

mass of bullet, m = 0.5 kg

initial velocity of bullet is u.

mass of pendulum, M = 9.50 kg

height raised, h = 1.28 m

The kinetic energy of the bullet and pendulum system is equal to the potential energy of the system

Let vf be the final velocity after the impact

KE = PE

1/2 (M + m) Vf^2 = (M + m) x g x h

vf^2 = 2 x 9.8 x 1.28 = 25.088

vf = 5 m/s

Thus, the final speed of the combined mass system after the impact s 5 m/s.  

Final answer:

The final speed of the combined masses after the collision can be found using the conservation of momentum and mechanical energy principles, resulting in a final speed of approximately 1.20 m/s.

Explanation:

The final speed Vf of the combined masses after the collision can be calculated using the principle of conservation of momentum and conservation of mechanical energy.

By equating the initial momentum of the system before the collision to the final momentum of the system after the collision, you can determine the final speed Vf.

In this case, the final speed Vf is approximately 1.20 m/s.

To create a gauge pressure of 62.5 cm H₂O on an area of 51 cm², you must exert a force of approximately 31.27 newtons on the balloon. This is obtained by converting the pressure to pascals, the area to square meters and using the formula Pressure = Force/Area.

To find the force you must exert on the balloon to create a gauge pressure of 62.5 cm H₂O when squeezing an area of 51 cm², you can use the relationship between pressure, force, and area given by the formula Pressure (P) = Force (F) / Area (A). First, convert the gauge pressure from cm H₂O to pascals (Pa) since 1 cm H₂O is approximately 98.0665 Pa. Then multiply the converted pressure by the area in m² to find the force in newtons (N).

Step 1: Convert cm H₂O to pascals (Pa):
62.5 cm H₂O * 98.0665 Pa/cm H₂O = 6131.65 Pa

Step 2: Convert the area from cm²to m² (since there are 10,000 cm2 in 1 m²):
51 cm² / 10,000 = 0.0051 m²

Step 3: Calculate the force:
Force (F) = Pressure (P) * Area (A)
F = 6131.65 Pa * 0.0051 m² = 31.27 N

Therefore, you must exert a force of approximately 31.27 newtons on the balloon.

Answer:

C. The bullet will hit the monkey because both the monkey and the bullet are falling downward at the same rate due to gravity.

Explanation:

As we know that acceleration due to gravity is equal g m/s².The acceleration due to gravity is in downward direction always and the numerical value is equal to g =10 m/s².

Both the bullet and money moving downward with constant acceleration that is why bullet will hit the money.

Therefore the answer is C.

Explanation:

Given that,

Distance between two long wires, d = 16 cm = 0.16 m

Current in one wire, [tex]I_1=2.9\ A[/tex]

Current in wire 2, [tex]I_2=5.7\ A[/tex]    

The magnetic force per unit length of one wire on the other is given by the following expression as :

[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi d}[/tex]

[tex]\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.9\times 5.7}{2\pi \times 0.16}[/tex]

[tex]\dfrac{F}{l}=2.06\times 10^{-5}\ N/m[/tex]

The current is flowing in opposite direction, the magnetic force acting on it is repulsive. Hence, this is the required solution.

Answer:

605447.7066 kgm²/s

Explanation:

[tex]m_1[/tex] = Mass of sphere = 10000 kg

[tex]m_2[/tex] = Mass of rod = 10 kg

r = Radius of sphere = 2 m

l = Length of antenna = 3 m

Angular speed

[tex]\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s[/tex]

Angular momentum is given by

[tex]L=I\omega[/tex]

Moment of inertia of the satellite is

[tex]I_s=\frac{2}{5}m_1r^2[/tex]

Moment of antenna of the satellite is

[tex]I_a=\frac{1}{3}m_2l^2[/tex]

The angular momentum of the system is

[tex]L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s[/tex]

The angular momentum of the satellite is 605447.7066 kgm²/s

The angular momentum of a rotating satellite can be calculated by considering the moment of inertia of its components, including the main body and antennas. The moment of inertia is calculated based on the mass and distribution of mass in each component. The total angular momentum is the sum of the angular momentum of the main body and the angular momentum of the antennas.

The angular momentum of a rotating object can be calculated by multiplying its moment of inertia with its angular velocity. In this case, the satellite consists of a main body in the shape of a sphere and two antennas projecting out from the center of mass. The moment of inertia can be calculated by considering the mass and distribution of mass in each component of the satellite. The angular momentum is the sum of the angular momentum of the main body and the angular momentum of the antennas.

For the main body of the satellite, the moment of inertia can be calculated as: I = (2/5)×m × r², where m is the mass and r is the radius of the sphere. For the antennas, the moment of inertia can be calculated as: I = (1/12) × m × L², where m is the mass and L is the length of each antenna.

Finally, the total angular momentum can be calculated by summing the angular momentum of the main body and the angular momentum of the antennas: L_total = L_main body + 2× L_antennas, since there are two antennas.

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Answer:

 [tex]c_{e1}[/tex] = 128.3 J / kg ° C

Explanation:

In this exercise we will use that the expression for heat is

    Q = m [tex]c_{e}[/tex] ΔT

As they indicate that there are no losses with the medium, the heat transferred by the tungsten is equal to the heat absorbed by the water plus the calorimeter

    Q assigned = QAbsorbed

    Q hot = Q cold + Q calorimeter

The mass of tungsten (m₁ = 69.12 10⁻³ kg) with an initial temperature (T₁ = 98.93°C),

The mass of water (m₂ = 85.45 10⁻³ kg) at a temperature (T₂ = 23.82°C),

a calorimeter constant (C = 1.56 J/ °C)

m₁ [tex]c_{e1}[/tex] (T₁ - [tex]T_{f}[/tex]) = (m₂ [tex]c_{e2}[/tex] + C) ([tex]T_{f}[/tex] - T₂)  

[tex]c_{e1}[/tex]= (m₂ ce2 + C) ([tex]T_{f}[/tex]-T₀) / (m₁ (T₁-[tex]T_{f}[/tex])  

[tex]c_{e1}[/tex] = (85.45 10-3 4186 + 1.56) (25.63 - 23.82) / (69.12 10-3 (98.93 - 25.63))  

[tex]c_{e1}[/tex] = (357.69 + 1.56) 1.81 / (69.12 10-3 73.3)  

[tex]c_{e1}[/tex] = 650.24 / 5.0665  

[tex]c_{e1}[/tex] = 128.3 J / kg ° C

Answer:

(a) 108

(b) 110.500 kW

(c) 920.84 A

Solution:

As per the question:

Voltage at primary, [tex]V_{p} = 120\ V[/tex]          (rms voltage)

Voltage at secondary, [tex]V_{s} = 13000\ V[/tex]  (rms voltage)

Current in the secondary, [tex]I_{s} = 8.50\ mA[/tex]  

Now,

(a) The ratio of secondary to primary turns is given by the relation:

[tex]\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}[/tex]

where

[tex]N_{p}[/tex] = No. of turns in primary

[tex]N_{s}[/tex] = No. of turns in secondary

[tex]\frac{N_{s}}{N_{p}} = \frac{13000}{120}[/tex] ≈ 108

(b) The power supplied to the line is given by:

Power, P = [tex]V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW[/tex]

(c) The current rating that the fuse should have is given by:

[tex]\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}[/tex]

[tex]\frac{13000}{120} = \frac{I_{p}}{8.50}[/tex]

[tex]I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A[/tex]

Answer:

a. 0 kgm/s

b. 0 kgm/s

c. 66 kgm/s

d. -66 kgm/s

e. 0 kgm/s

f. -27.05 m/s

g. 173.68 N

h. 12.58 m/s

i. 0.772 m

j. 14487 J

Explanation:

150 g = 0.15 kg

a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.

b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.

c. After the bullet is fired, the momentum is:

0.15*440 = 66 kgm/s

d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s

e. 0 according to law of momentum conservation.

f. Velocity of the rifle is its momentum divided by mass

v = -66 / 2.44 = -27.05 m/s

g. The average force would be the momentum divided by the time

f = -66 / 0.38 = 173.68 N

h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is

66 / 5.25 = 12.58 m/s

i. The normal force and also friction force due to sliding is

[tex]F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N[/tex]

According to law of energy conservation, the initial kinetic energy will soon be transformed to work done by this friction force along a distance d:

[tex]W = K_e[/tex]

[tex]dF_f = 0.5Mv_0^2 [/tex]

[tex]d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m[/tex]

j.Kinetic energy of the bullet before the impact:

[tex]K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J[/tex]

Kinetic energy of the block-bullet system after the impact:

[tex]K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J[/tex]

So 14520 - 33 = 14487 J was lost during the lodging process.

Answer:

2.27m/s

Explanation:

The coefficient of sliding friction = 0.38, the incline angle = 60o

Coefficient of sliding friction = frictional force/ force of normal (fn)

The normal force = mgcos60 where m is the mass of the body and g is acceleration due to gravity

Frictional force = coefficient of friction*fn = 0.38*9.81*m*cos 60 = 1.8639m

Force tending to push the body down the incline plane = mgsin60 = 8.496m

Net force on the body = force pushing downward - Frictional force

Net force = 8.496m - 1.8639m = 6.632m

Kinetic energy of the body = 1/2 mv^2 = work done by the net force pushing the body down = net force * distance travelled

1/2mv^2 = 6.632m * 0.387in meters

Cancel mass on both side leaves 1/2v^2 = 6.632*0.387

V^2 = 2.566584*2 = 5.133

V = √5.133 = 2.27m/s

The speed of the block after it has traveled a distance of 38.7 cm is 1.2 m/s.

The speed of the block is calculated by applying the following formula.

Using work energy theorem;

The work done by the force of friction is equal to the change in kinetic energy of the box.

W = K.E

μmgd cosθ  = ¹/₂mv²

μgdcosθ = ¹/₂v²

v² = (2gd cosθ)μ

v = √ (2gd cosθμ)

where;

v = √ (2gd cosθμ)

v = √ (2 x 9.8 x 0.387  x cos(60) x 0.38)

v = 1.2 m/s

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Answer:

4.5 x 10^6 miles

Calculations can be viewed on the snapshot attached to this reply.

Thanks

Answer:

The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Explanation:

Hi there!

The total momentum is calculated as the sum of the momenta of the pieces.

The momentum of each piece is calculated as follows:

p = m · v

Where:

p = momentum.

m =  mass.

v = velocity.

The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:

p = (m · v, 0)

p = (0.200 kg · 82.0 m/s, 0)

p = (16.4 kg m/s, 0)

The 300 g-piece flies along the y-axis. Its momentum vector will be:

p =(0, m · v)

p = (0, 0.300 kg · 45.0 m/s)

p = (0, 13.5 kg m/s)

The total momentum is the sum of each momentum:

Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)

Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)

Total momentum = (16.4 kg m/s, 13.5 kg m/s)

The magnitude of the total momentum is calculated as follows:

[tex]|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s[/tex]

The direction of the momentum vector is calculated using trigonometry:

cos θ = px/p

Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.

cos θ = 16.4 kg m/s / 21.2 kg m/s

θ = 39.3  (39.5° if we do not round the magnitude of the total momentum)

Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

Mathematical expression for the Newton's second law of motion is given as:

[tex]F=\frac{dp}{dt}[/tex] ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

[tex]p=m.v[/tex]

Now, equation (1) becomes:

[tex]F=\frac{d(m.v)}{dt}[/tex]

∵mass is constant at speeds v << c (speed of light)

[tex]\therefore F=m.\frac{dv}{dt}[/tex]

and, [tex]\frac{dv}{dt} =a[/tex]

where: a = acceleration

[tex]\Rightarrow F=m.a[/tex]

also

[tex]F\propto \frac{1}{dt}[/tex]

so, more the time, lesser the force.

& Impulse:

[tex]I=F.dt[/tex]

[tex]I=m.a.dt[/tex]

[tex]I=m.\frac{dv}{dt}.dt[/tex]

[tex]I=m.dv=dp[/tex]

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

Final answer:

According to the impulse-momentum theorem, the air mattress exerts the same impulse as the hard ground on the high-jumper, but with a smaller net average force because the stopping time is extended on the air mattress.

Explanation:

The key to understanding this scenario lies in the impulse-momentum theorem, which states that the change in momentum (impulse) of an object is equal to the average net external force applied to it, times the duration of time this force acts upon the object.

In both cases, landing on the hard ground and landing on an air mattress, the change in momentum of the high-jumper is the same, since she comes to rest from the same initial velocity in both scenarios. Thus, the overall impulse received by the high-jumper must be the same in both situations. However, the difference lies in the time over which the stopping force is applied.

When landing on the air mattress, the stopping time is longer than on the hard ground. According to the impulse-momentum theorem (Ap = Fnet At), since the impulse (Ap) is the same but the time (t) is greater when landing on the air mattress, the average net force (Fnet) must be smaller in comparison with landing on the hard ground. Therefore, the correct statement is that the air mattress exerts the same impulse but a smaller net average force than the hard ground.

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

Let the mass of each ball be m kg

v[tex]_{1}[/tex] be the velocity of ball A along positive x axis

v[tex]_{2}[/tex] be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v[tex]_{1}[/tex]

∴  v[tex]_{1}[/tex] = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v[tex]_{2}[/tex]

2 = u +  v[tex]_{2}[/tex] → equation 1

∴ relative velocity of approach = relative velocity of separation

-2 =  v[tex]_{2}[/tex] - u → equation 2

By adding both equations 1 and 2 we get

v[tex]_{2}[/tex] = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

To solve this problem it is necessary to apply the concepts related to the concepts to continuity.

From the continuity equation we know that

[tex]A_1V_1 = A_2 V_2[/tex]

Where,

A = Cross-sectional Area

V = Velocity

In a circle the area is given by

[tex]A_1 = \pi r_1^2\\A_2 = \pi r_2^2[/tex]

Therefore replacing with our values we have that

[tex]A_1 = \pi r_1^2\\A_1 = \pi 1.8^2\\A_1 = 10.278m^2[/tex]

While the area two is defined as

[tex]A_2 = \pi r_2^2\\A_2 = \pi 0.6^2\\A_2 = 1.13m^2[/tex]

Applying the continuity equation we have to

[tex]A_1V_1 = A_2 V_2[/tex]

[tex](10.278)(3)= (1.13) V_2[/tex]

[tex]V_2 = 27.28m/s[/tex]

Therefore the speed of water flow in the test section is 27.28m/s

Answer:

Explanation:

Given that,

Radius of the wheel, r = 20 cm = 0.2 m

Initial speed of the wheel, [tex]\omega_i=120\ rpm=753.98\ rad/s[/tex]

Displacement, [tex]\theta=90\ rev=565.48\ rad[/tex]

To find,

The angular acceleration and the distance covered by the car.

Solution,

Let [tex]\alpha[/tex] is the angular acceleration of the car. Using equation of rotational kinematics as :

[tex]\theta=\omega_i t+\dfrac{1}{2}\alpha t^2[/tex]

[tex]565.48=753.98\times 60+\dfrac{1}{2}\alpha (60)^2[/tex]

[tex]\alpha =-24.81\ rad/s^2[/tex]

Let t is the time taken by the car before coming to rest.

[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }[/tex]

[tex]t=\dfrac{0-753.98}{-24.81}[/tex]

t = 30.39 seconds

Let v is the linear velocity of the car. So,

[tex]v=r\times \omega_i[/tex]

[tex]v=0.2\times 753.98[/tex]

v = 150.79 m/s

Let d is the distance covered by the car. It can be calculated as :

[tex]d=v\times t[/tex]

[tex]d=150.79\ m/s\times 30.39\ s[/tex]

d = 4582.5 meters

or

d = 4.58 km

q3 should be placed at x = + 6.9 cm to make the potential energy of the system equal to zero.

To calculate the potential energy of the system of three charges, we use the formula:

PE = k * q1 * q2 / r12 + k * q2 * q3 / r23 + k * q3 * q1 / r31

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q1, q2, and q3 are the charges of the three point charges (4.10 nC, -2.90 nC, and 2.00 nC, respectively)

r12, r23, and r31 are the distances between the three point charges (20.0 cm, 9.0 cm, and 11.0 cm, respectively)

Plugging in the values, we get:

PE = (8.98755 x 10^9 N·m^2/C^2) * (4.10 x 10^-9 C) * (-2.90 x 10^-9 C) / 0.20 m + (8.98755 x 10^9 N·m^2/C^2) * (-2.90 x 10^-9 C) * (2.00 x 10^-9 C) / 0.09 m + (8.98755 x 10^9 N·m^2/C^2) * (2.00 x 10^-9 C) * (4.10 x 10^-9 C) / 0.11 m

PE = 1.19 x 10^-6 J

Therefore, the potential energy of the system of three charges if q3 is placed at x = + 11.0 cm is 1.19 x 10^-6 J.

To make the potential energy of the system equal to zero, we need to place q3 at a point where the forces from q1 and q2 cancel each other out. This means that the vectors representing the forces must be equal and in opposite directions.

The force from q1 on q3 is given by:

F13 = k * q1 * q3 / r31^2

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q1 is the charge of q1 (4.10 nC)

q3 is the charge of q3 (2.00 nC)

r31 is the distance between q1 and q3

The force from q2 on q3 is given by:

F23 = k * q2 * q3 / r23^2

where:

k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)

q2 is the charge of q2 (-2.90 nC)

q3 is the charge of q3 (2.00 nC)

r23 is the distance between q2 and q3

For the forces to cancel each other out, we need:

F13 = -F23

Substituting in the equations for the forces, we get:

k * q1 * q3 / r31^2 = -k * q2 * q3 / r23^2

Solving for r31, we get:

r31 = r23 * sqrt(q2/q1)

Substituting in the values, we get:

r31 = 0.09 m * sqrt(-2.90 nC / 4.10 nC)

r31 = 0.069 m

Therefore, q3 should be placed at x = + 6.9 cm to make the potential energy of the system equal to zero.

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Answer:

The cheerleader does a total work of 8,532 J (8,541 J without any intermediate rounding)

Explanation:

Hi there!

The equation of work is the following:

W = F · d

Where:

W = work.

F = applied force.

d = distance.

The cheerleader´s partner is on the ground due to the force of gravity acting on her, also known as weight. To lift the partner at a constant velocity, the cheerleader has to suppress this force applying on the partner a force that is equal to the partner´s weight but in opposite direction.

The weight of the partner is calculated as follows:

Weight = m · g

Where m is the mass of the partner and g is the acceleration due to gravity.

Let´s calculate the weight of the partner:

Weight = 43.8 kg · 9.8 m/s²

Weight = 429 N

]Then, the work done in one lift by the cheerleader is the following:

W = F · d

W = 429 N · 0.737 = 316 J

If the cheerleader does 27 lifts, the work done will be (316 J · 27) 8,532 J

The work done by the cheerleader to lift his partner 27 times is calculated as the product of force (which itself is the product of the partner's mass and gravity) and distance, hence resulting in a total work done of 8570.61 Joules.

In Physics, energy exerted to move an object is often calculated as work done. Work is defined as the product of the force applied to an object and the distance over which it is applied, and its unit of measurement is Joules (J). Regarding the given problem, the work done by the cheerleader to lift his partner is calculated using the formula: Work = force x distance, where force is derived from the product of the partner's mass and gravity (Force = mass x gravity). So, the work done for one lift would be calculated as follows:

Work(std.) = Force * distance = mass * gravity * distance = 43.8 kg * 9.8 m/s ² * 0.737 m = 317.43 J.

To find out the total work done after 27 lifts, simply multiply this result by 27: Total Work = 27 * Work(std.) = 27 * 317.43 J = 8570.61 J. Thus, the cheerleader does 8570.61 Joules of work by lifting his partner 27 times.

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To find the heat required to melt the ice on the windshield, calculate the mass of the ice using its thickness, area, and density, then use the formula for heat transfer to find the heat required using the latent heat of fusion for ice.

In this Physics problem, we're trying to calculate the heat needed to melt the -4.0 °C ice from a windshield. The heat required to melt ice is given by the formula Q = m * Lf, where Q is the heat, m is the mass of the substance (ice in this case), and Lf is the latent heat of fusion, which is the amount of heat necessary to change the state of the substance from solid to liquid at its melting point.

To calculate the mass of the ice, we use the density (ρ) given and the volume of the ice, which can be found by multiplying the thickness of the ice by the area of the windshield and converting cm to m. Volume of the ice would be Area * thickness = 1.0 m^2 * 0.007 m (from 0.7 cm) = 0.007 m^3. Plug this into ρ = m/V to get the mass m = ρ * V = 917 kg/m^3 * 0.007 m^3 = 6.42 kg.

Finally, we use Q = m * Lf which is Q = 6.42 kg * 334 kJ/kg (given that the heat of fusion for water is 334 kJ/kg), we get Q = 2143 kJ required to melt all the ice.

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Answer:

0.15694 m

Explanation:

m = Mass of block = [tex]2.16\times 10^{-2}\ kg[/tex]

v = Velocity of block = 11.2 m/s

k = Spring constant = 110 N/m

Here the kinetic energy of the fall and spring are conserved

[tex]\frac{1}{2}mv^2=\frac{1}{2}kA^2\\\Rightarrow mv^2=kA^2\\\Rightarrow A=\sqrt{\frac{mv^2}{k}}\\\Rightarrow A=\sqrt{\frac{2.16\times 10^{-2}\times 11.2^2}{110}}\\\Rightarrow A=0.15694\ m[/tex]

The amplitude of the resulting simple harmonic motion is 0.15694 m

Answer:

31.66 m/s

Explanation:

mass of player, M = 75 kg

mass of ball, m = 0.45 kg

initial velocity of player, U = + 4 m/s

initial velocity of ball, u = - 24 m/s

Let the final speed of player is V and the ball is v.

use conservation of momentum

Momentum before collision = momentum after collision

75 x 4 - 0.45 x 24 = 75 x V + 0.45 x v

289.2 = 75 V + 0.45 v    .... (1)

As the collision is perfectly elastic, coefficient of restitution,e = 1

So, [tex]e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}[/tex]

V - v = u - U

V - v = -24 - 4 = - 28

V = v - 28, put this value in equation (1), we get

289.2 = 75 (v - 28) + 045 v

289.2 = 75 v - 2100 + 0.45 v

2389.2 = 75.45 v

v = 31.66 m/s

Thus, the velocity of ball after collision is 31.66 m/.

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that

[tex]\sum F = ma[/tex]

There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is

[tex]F - F_f = ma[/tex]

To find the required force then,

[tex]F=F_f+ma[/tex]

By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say

[tex]F = \mu mg +ma[/tex]

[tex]F = 0.35*50*0.8+50*1.2[/tex]

[tex]F=(171.5N)+(50Kg)(1.2m/s^2)[/tex]

[tex]F=231.5N[/tex]

[tex]F\approx 230N[/tex]

Therefore the horizontal force applied on the block is B) 230N

To find the horizontal force necessary to accelerate the box at 1.2 m/s^2, we consider the frictional forces. The maximum static friction can be calculated using the coefficient of static friction and the weight of the box. The horizontal force required is equal to the maximum static friction.

To find the horizontal force required to accelerate the 50-kg box at 1.2 m/s^2, we need to consider the frictional forces acting on the box. First, we calculate the maximum static friction using the formula:

Maximum static friction = coefficient of static friction × normal force

Normal force = mass × gravity

Maximum static friction = 0.65 × (50 kg × 9.8 m/s^2)

Horizontal force = maximum static friction = 0.65 × (50 kg × 9.8 m/s^2)

So, the horizontal force needed is approximately 318.5 N.

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Answer:

The speed of the car was 28 m/s.

Explanation:

Hi there!

The initial kinetic energy of the car, KE, is equal to the negative work, W, done by friction to bring the car to stop. Let´s write the work-energy theorem:

W = ΔKE = final kinetic energy - initial kinetic energy

In this case, the final kinetic energy is zero, then:

W = - initial kinetic energy

Since the work done by friction is negative (the work is done in opposite direction to the movement of the car), then:

Wfr = initial kinetic energy

The work done by friction is calculated as follows:

Wfr = Fr · d

Where:

Fr = friction force.

d = distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction

Since the only vertical forces acting on the car are the weight of the car and the normal force, and the car is not being accelerated in the vertical direction, the normal force has to be equal to the weight of the car (with opposite sign).

Then the friction force can be written as follows:

Fr = m · g · μ

Where:

m = mass of the car.

g =  acceleration due to gravity (9.8 m/s²)

The work done by friction will be:

W = m · g · μ · d

The equation of kinetic energy is the following:

KE =  1/2 · m · v²

Where

m = mass of the car.

v = speed.

Then:

W = KE

m · g · μ · d = 1/2 · m · v²

2 · g · μ · d = v²

2 · 9.8 m/s² · 0.40 · 98 m = v²

v = 28 m/s

The speed of the car was 28 m/s.

In order to solve this problem, it is necessary to apply the concepts related to the Pressure according to the Force and the Area as well as to the pressure depending on the density, gravity and height.

In the first instance we know that the pressure can be defined as

[tex]P = \frac{F}{A}[/tex]

Where

F= Force

A =  Mass

In the second instance the pressure can also be defined as

[tex]P = \rho gh[/tex]

Where,

[tex]\rho=[/tex]Density of Fluid at this case Water

g = Gravitational Acceleration

h = Height

If we develop the problem to find the pressure then,

[tex]P = \frac{F}{A}[/tex]

[tex]P = \frac{F}{\pi r^2}[/tex]

[tex]P = \frac{2*10^6}{\pi (0.1)^2}[/tex]

[tex]P = 6.37*10^{7} Pa[/tex]

Through the second equation we can find the depth to which it can be submerged,

[tex]P = \rho gh[/tex]

Re-arrange to find h

[tex]h = \frac{P}{\rho g}[/tex]

[tex]h = \frac{6.37*10^{7}}{1000*9.8}[/tex]

[tex]h = 6499.4m[/tex]

To solve this problem it is necessary to apply the concepts related to the Pressure which describes the amount of Force made on an area unit.

Its mathematical expression can be defined as

[tex]P = \frac{F}{A}[/tex]

Where,

F= Force

A = Area

Our values are given as

[tex]d = 0.987*10^{-2}m[/tex]

With the diameter we can find the Area of the circular heel, that is

[tex]A = \pi (\frac{0.987*10^{-2}}{2})^2[/tex]

[tex]A = 7.6511*10^{-5}m^2[/tex]

To find the force by weight we need to apply the Newton's second Law

[tex]F = mg[/tex]

[tex]F = 53*9.8[/tex]

[tex]F = 519.4N[/tex]

Finally the pressure would be

[tex]P = \frac{F}{A}[/tex]

[tex]P = \frac{519.4}{7.6511*10^{-5}}[/tex]

[tex]P = 6788566.35Pa[/tex]

[tex]P = 6.788MPa[/tex]

Therefore the pressure that she exerts on the carpet sample is 6.788Mpa

Answer:m=86.36 kg

Explanation:

Given

mass of tackler [tex]m_T=100 kg[/tex]

initial velocity of Tackler [tex]u_t=4.1 m/s[/tex]

Final velocity of combined system [tex]v=2.2 m/s[/tex]

Let m be the mass of receiver

conserving momentum

[tex]m\times 0+m_t\times u_t=(m+m_t)v[/tex]

[tex]0+100\times 4.1=(100+m)\cdot 2.2[/tex]

[tex]410=(100+m)\cdot 2.2[/tex]

[tex]100+m=186.36[/tex]

[tex]m=86.36 kg[/tex]

To solve this problem it is necessary to apply the concepts related to energy as a function of voltage, load and force, and the definition of Force given by Newton in his second law.

By definition we know that force is equal to

F= ma

Where,

m = mass (at this case of an electron)

a = Acceleration

But we also know that the Energy of an electric object is given by two similar definitions.

[tex]1) E= \frac{F}{q}[/tex]

Where,

F= Force

q = Charge of proton/electron

[tex]2) E = \frac{V}{d}[/tex]

V = Voltage

d = Distance

Equating and rearrange for F,

[tex]\frac{F}{q} = \frac{V}{d}[/tex]

[tex]F = \frac{Vq}{d}[/tex]

The two concepts of force can be related to each other, then

[tex]ma = \frac{Vq}{d}[/tex]

Acceleration would be,

[tex]a = \frac{Vd}{dm}[/tex]

Replacing with our values we have that the acceleration is

[tex]a = \frac{Vq}{dm}[/tex]

[tex]a = \frac{(170)(1.6*10^{-19})}{(2*10^{-2})(9.1*10^{-31})}[/tex]

[tex]a = 1.49*10^{15}m/s^2[/tex]

Now through the cinematic equations of motion we know that,

[tex]V_f^2-V_i^2 = 2ax[/tex]

Where,

[tex]V_f =[/tex] Final velocity

[tex]V_i =[/tex] Initial velocity

a = Acceleration

x = Displacement

Re-arrange to find v_f,

[tex]v_f = \sqrt{v_i^2+2ax}[/tex]

[tex]v_f = \sqrt{2.5*10^5+2*(1.49*10^{15})(0.1*10^{-2})}[/tex]

[tex]v_f = 1.726*10^6 m/s[/tex]

Therefore the electron's speed when it is 0.1 cm from the negative plate is [tex]1.726*10^6 m/s[/tex]

Speed of the electron enters in the gape of two large aluminum plates,  when it is 0.1 cm from the negative plate is,

[tex]v_f=1.726\times10^6\rm m/s[/tex]

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

The electric force due to the electric field can be given as,

[tex]F=\dfrac{Vq}{d}[/tex]

Here, (V) is the potential difference, (q) is the charge on the body and (d) is the distance.

The charge on a electron is [tex]1.6\times10^{-19}[/tex] C. As the distance between the two plates is 2 meters and the value of potential difference is 170 V. Thus, put the values in the above formula as,

[tex]F=\dfrac{170\times(1.6\times10^{-19})}{0.02}[/tex]

As the force on a body is the product of mass times acceleration and the mass of the electron is [tex]9.1\times10^{-31}[/tex] kg. Thus, the above equation can be rewrite as,

[tex](9.1\times10^{-31})a=\dfrac{170\times(1.6\times10^{-19})}{0.02}[/tex]

[tex]a=1.49\times10^{15}\rm m/s^2[/tex]

The initial speed of the electron is [tex]2.5\times10^5[/tex] m/s and the acceleration is [tex]1.49\times10^{15}\rm m/s^2[/tex]. Thus from the third equation of motion, the final velocity can be given as,

[tex]v_f=\sqrt{2.5\times10^5+2(1.49\times10^{15})(0.1\times10^{-2})}\\v_f=1.726\times10^6\rm m/s[/tex]

Thus, the speed of the electron,  when it is 0.1 cm from the negative plate is,

[tex]v_f=1.726\times10^6\rm m/s[/tex]

Learn more about electric field here;

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Answer:

  W = 661.6 N

Explanation:

The weight of a body is the force of attraction of the plant on the body, so we must use the law of gravitational attraction

       F = G m M / r²

Where G is the gravitational attraction constant that values ​​6.67 10-11 N m² / kg², M is the mass of the planet and r is the distance from the center of the planet.

Let's look for the mass of the planet, for this we write Newton's second law for the landing craft

     F = m a

Acceleration is centripetal a = v² / r

     G m M / r² = m (v² / r)

The ship rotates rapidly (constant velocity module), let's use uniform kinematic relationships

    v = d / t

The distance of a circle is

    d = 2π r

    v = 2π r / t

We replace

     G m M / r² = m (4π² r² / t² r)

    G M = 4 π² r³ / t²

    M = 4π² r³ / G t²

The measured distance r from the center of the plant is

     r = R orbit + R planet

     r = 5.90 10⁵ + ½ 9.80 10⁶

     r = 5.49 10⁶ m

    M = 4 π² (5.49 10⁶)³ / (6.67 10⁻¹¹ (5.900 10³)²)

    M = 6,532 10²¹ / 2,321 10⁺³

    M = 2.814 10²⁴ kg

With this data we calculate the astronaut's weight

     W = (G M / R²) m

     W = (6.67 10⁻¹¹ 2,816 10²⁴ /(4.90 10⁶)2)   84.6

     W = 7.82  84.6

     W = 661.57 N

To calculate the weight of the astronaut on the surface of the planet, we first need to find out the gravitational acceleration at the planet's surface using the parameters of the landing craft's orbit. After we have the value of gravitational acceleration, we multiply it by the mass of the astronaut.

To answer your question, we first need to calculate the gravitational acceleration at the surface of the planet, denoted by g. We can use the orbital parameters of the landing craft to find g using the following formula where G represents the universal gravitation constant, M is the mass of the planet, and r is the radius of the planet:

g = GM/r^2

After we solve this, we can then calculate the weight of the astronaut on the surface using W = mg, where m is the mass of the astronaut.

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The number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.

Answer: Option B

Explanation:

As a result of photoelectric effect, electrons are emitted by the light incident on a metal surface. The emitted electrons count and its kinetic energy can measure as the function of light intensity and frequency. Like physicists, at the 20th century beginning, it should be expected that the light wave's energy (its intensity) will be transformed into the kinetic energy of emitted electrons.

In addition, the electrons count emitting from metal must vary with light wave frequency. This frequency relationship was expected because the electric field oscillates due to the light wave and the metal electrons react to different frequencies. In other words, the number of electrons emitted was expected to be frequency dependent and their kinetic energy should be dependent on the intensity (constant wavelength) of light.

Thus, the maximum in kinetic energy of electrons emitted increases with increase in light's frequency and is experimentally independent of light intensity. So, the number of emitted electrons is proportionate to the intensity of the incident light.

Final answer:

If the intensity of the incident light on a metal surface in a photoelectric effect experiment is increased, while keeping frequency constant, the number of electrons emitted per second increases. This happens because a higher intensity means more photons are available to eject electrons. Neither the work function of the metal, the maximum speed of the emitted electrons, nor the stopping potential are affected by changes in light intensity.

Explanation:

The question relates to the effect of increasing the intensity of light in a photoelectric effect experiment while keeping the frequency of the incident light and the temperature of the metal constant. In this setup:

It is important to note that:

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:  

[tex]p_i = p_f[/tex]  

[tex]m_1u_1 + m_2v_2 = (m_1 + m_2)v[/tex]

[tex]v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)} [/tex]

[tex]v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s[/tex]  

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:  

[tex]E(i) = E(f)[/tex]  

[tex]KE(i) + PE(i) = KE(f) + PE(f)[/tex]  

[tex]0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)[/tex]  

[tex]v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}[/tex]

Here, initial velocity is the final velocity from the first stage. Therefore:  

[tex]v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s[/tex]

3. Use conservation of momentum to find the combined speed of Gayle and her brother.  

Given:

Initial velocity of Gayle and sled is, [tex]u_1(i)=10.5[/tex] m/s

Initial velocity of her brother is, [tex]u_2(i)=0[/tex] m/s

Mass of Gayle and sled is, [tex]m_1=55.0[/tex] kg

Mass of her brother is, [tex]m_2=30.0[/tex] kg

Final combined velocity is given as:

[tex]v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}[/tex]  

[tex]v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79 [/tex] m/s  

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:  

[tex]E(i) = E(f)[/tex]  

[tex]KE(i) + PE(i) = KE(f) + PE(f)[/tex]  

[tex]0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)[/tex]  

[tex]v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s[/tex]

Using principles of kinetic and potential energy, and assuming ideal conditions devoid of friction and air resistance, Gayle and her brother's final speed would be equal to the square root of twice the product of acceleration due to gravity and the total height of the hill.

This question relates to the principles of kinetic energy and potential energy, as well as conservation of energy. When Gayle initially sleds down the hill, she converts potential energy (mgh) into kinetic energy (1/2 mv^2). Taking 'g' as acceleration due to gravity (9.8 m/s^2) and 'h' as the vertical height she descends (5.0 m), the speed she attains at the time her brother joins (v') can be calculated using √(2gh).

The potential energy at that point is their combined mass times gravity and their height, which is transformed to kinetic energy at the bottom. If 'H' is the total height of the hill (15.0 m), their final combined speed is given by √(2gH). Ignoring resistive forces, they maintain this speed for the entire ride down the hill.

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Answer:

Induced EMF,[tex]\epsilon=0.0143\ volts[/tex]

Explanation:

Given that,

Radius of the circular loop, r = 5 cm = 0.05 m

Time, t = 0.0548 s

Initial magnetic field, [tex]B_i=200\ mT=0.2\ T[/tex]

Final magnetic field, [tex]B_f=300\ mT=0.3\ T[/tex]

The expression for the induced emf is given by :

[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]

[tex]\phi[/tex] = magnetic flux

[tex]\epsilon=\dfrac{d(BA)}{dt}[/tex]

[tex]\epsilon=A\dfrac{d(B)}{dt}[/tex]

[tex]\epsilon=A\dfrac{B_f-B_i}{t}[/tex]

[tex]\epsilon=\pi (0.05)^2\times \dfrac{0.3-0.2}{0.0548}[/tex]

[tex]\epsilon=0.0143\ volts[/tex]

So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.

Final answer:

The magnitude of the induced emf in the loop is -0.0184 V.

Explanation:

The magnitude of the emf induced in a single-turn circular loop of wire can be calculated using Faraday's law of electromagnetic induction. According to the law, the magnitude of the induced emf is equal to the rate of change of magnetic flux through the loop. The formula for calculating the induced emf is:

emf = -N * ΔB/Δt

Where emf is the induced electromotive force, N is the number of turns in the loop, ΔB is the change in magnetic field strength, and Δt is the change in time. Plugging in the given values, we have:

emf = -1 * (300 - 200) * 10^-3 / 0.0548

emf = -0.0184 V