A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the


Answer 1


v = 0.85 m/s


Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :



v=\sqrt{(0.025* 0.29)/(0.01)}

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

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A baseball player throws a ball into the stands at 15.0 m/s and at an angle 45.0° above the horizontal. On its way down, the ball is caught by a spectator 4.10 m above the point where the ball was thrown. How much time did it take for the ball to reach the fan in the stands?



Time = 1.61 seconds


Using the equation displacement of a trajectory motion in the y plane

Y = u t sin ů - ½gt²....equation 1 where

Y= vertical displacement =4.1

U = initial velocity = 15m/s

g = acc. Due to gravity = 10m/s

Ů = angle of trajectory = 45

t = time to reach fan on its way down

Sub into equ 1

4.1 = 15t sinů - ½ * 10t²

4.1 = 10.61t - 5t²

Solve using quadratic formula

t =[-B±( -B² -4AC)^½]/2A....equation 2

Where A = 5, B=10.61, C =4.1

Substitute A,B,C into equ2

t = (10.61±5.53)/10

t = 0.508seconds or 1.61seconds

Since it is on its way down t= 1.61 seconds

A stream of water emerging from a faucet narrows as fails. The cross-sectional area of the soutis As -6.40 cm. water comes out of the spout at a speed of 33.2 cm/s, and the waterfalls h = 7.05 cm before iting the bottom of sink What is the cross-sectional area of the water stream just before it is the sink? a. 0.162 cm3 b. 1.74 cm3c. 6.21cm3d. 0.943cm3



The area  of the water stream will be 1.74 cm^2


initial velocity of water u = 33.2 cm/s

initial area = 6.4 cm^2

height of fall = 7.05 cm

final area before hitting the sink = ?

as the water falls down the height, it accelerates under gravity; causing the speed to increase, and the area to decrease.

first we find the velocity before hitting the sink


v^(2) = u^(2)  + 2gh  -----Newton's equation of motion

where  v is the velocity of the water stream at the sink

u is the initial speed of the water at the spout

h is the height of fall

g is acceleration due to gravity, and it is positive downwards.

g = 981 cm/s^2

imputing relevant values, we have

v^(2) = 33.2^(2)  + (2 * 981 * 7.05)

v^(2) = 1102.24  + 13832.1 = 14934.34

v = √(14934.34) = 122.206 cm/s

according to continuity equation,

A1v1 = A2v2

where A1 is the initial area

V1 = initial velocity

A2 = final area

V2 = final velocity

6.4 x 33.2 = 122.206 x A2

212.48 = 122.206 x A2

A2 = 212.48 ÷ 122.206 ≅ 1.74 cm^2

Susan’s 10 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30° above the floor. The tension is a constant 30 N and the coefficient of friction is 0.20. Use work and energy to find Paul’s speed after being pulled 3.0 m.


Final answer:

In this problem, we have calculated the work done by Susan pulling her baby brother on a mat and the work done against friction. The net work done, which is the work done by Susan's pulling minus the work done against friction, is transformed into kinetic energy, giving us the baby's speed after being pulled 3m, which is approximately 1.95 m/s.


To answer this question, we first need to calculate the work done by Susan when she pulls the mat over the distance of 3.0 meters. The angle at which the rope is pulled does make a difference in this calculation. The force that is actually contributing to the work is the horizontal component of the tension, which can be determined by the equation Fh = F cos θ which equals 30N * cos30 = 25.98N.

The work done, W, is equal to this force multiplied by the displacement, so W = Fd = 25.98N * 3m = 77.94 Joules.

Next, we need to calculate the work done against friction. The force of friction is calculated as Ff = µN. Here N is the normal force, which is equal to the weight of the baby, so N = mg = 10kg * 9.8m/s² = 98N. The force of friction then is Ff = µN = 0.20 * 98N = 19.6N. The work done against friction is Wf = Ff * d = 19.6N * 3m = 58.8 Joules.

The net work done on the baby is the work done by Susan minus the work done against friction, so Wnet = W - Wf = 77.94J - 58.8J = 19.14 Joules. This net work is equal to the change in kinetic energy of the baby, ∆K, since Kinitial = 0 (Paul starts at rest), the work done is all transformed into final kinetic energy. So ∆K = 19.14J.

The kinetic energy of an object is given by the equation K = 1/2 mv², so we have 19.14J = 1/2 * 10kg * v². Solving for v gives us roughly v = 1.95 m/s. Therefore, the speed of the baby after being pulled 3 meters is approximately 1.95 m/s.

Learn more about Physics: Work and Energy here:



Final answer:

To determine Paul's speed, we must calculate the net work done on him using the work-energy theorem. This includes the work done by Susan and the work done against friction. Paul’s speed after being pulled 3.0 m is approximately 1.96 m/s.


Solving this problem involves understanding the work-energy theorem and forces. First, let's calculate the work done. The work done by the force Susan applies (W1) is the product of the tension (T), the distance (d), and the cosine of the angle (θ). W1 = T * d * cos(θ) = 30N * 3.0m * cos(30) = 77.94J.

Next, the work done against friction (W2) is the product of the frictional force and the distance, which is µmgd. Here, µ is the coefficient of friction (0.20), m (10kg) is the mass of the baby, g (9.8m/s2) is the acceleration due to gravity, and d is the distance (3.0 m). W2 = µmgd = 0.20 * 10kg * 9.8m/s2 * 3.0m = 58.8J.

According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. Therefore, the final kinetic energy (and thus the final speed) of Paul will be the initial kinetic energy plus the net work done on him. His initial speed is assumed to be zero, hence the initial kinetic energy is zero. The net work done on him is W = W1 - W2= 77.94J - 58.8J = 19.14J. Setting this equal to the final kinetic energy, (1/2)mv2, allows us to solve for the final speed, v = sqrt((2 * W)/m) = sqrt((2 * 19.14J)/10kg) = 1.96 m/s approximately.

Learn more about the Work-Energy Theorem here:



A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separation s could you produce in a typical atom by applying



s = 6.25 10⁻²² m


Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

                p = q s

               s = p / q

let's calculate

              s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

              s = 0.625 10⁻²¹ m

              s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.0 m. Determine the magnitude of the impulse delivered to the ball by the floor.


Given Information:

Mass of ball = m = 0.320 kg

Initial height = h₁ = 19 m

Final height = h₂ = 15 m

Required Information:

Impulse = I = ?


Impulse = 11.77 kg.m/s


We know that impulse is equal to change in momentum

I = Δp

I = p₁ - p₂

I = mv₁ - mv₂

I = m(v₁ - v₂)

Where m is the mass of ball, v₂ is the final velocity of the ball, and v₁ is the initial velocity of the ball.

So first we need to find the initial and final velocities of the ball

The relation between initial potential energy and final kinetic energy before the collision is given by

PE₁ = KE₂

mgh₁ = ½mv₂²

gh₁ = ½v₂²

v₂² = 2gh₁

v₂ = √2gh₁

v₂ = √2*9.8*19

v₂ = 19.3 m/s

The relation between initial kinetic energy and final potential energy after the collision is given by

KE₁ = PE₂

½mv₁² = mgh₂

½v₁² = gh₂

v₁² = 2gh₂

v₁ = √2gh₂

v₁ =√2*9.8*15

v₁ = 17.15 m/s

Finally, we can now find the magnitude of the impulse delivered to the ball by the floor.

I = 0.320(17.5 - (-19.3))

I = 11.77 kg.m/s


Imp = 11.666\,(kg\cdot m)/(s)


Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

(0.32\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot (19\,m) = (1)/(2)\cdot (0.320\,kg)\cdot v_(A)^(2)

v_(A) \approx 19.304\,(m)/(s)

After collision:

(1)/(2)\cdot (0.320\,kg)\cdot v_(B)^(2) = (0.32\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot (15\,m)

v_(B) \approx 17.153\,(m)/(s)

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

Imp = (0.32\,kg)\cdot [(17.153\,(m)/(s) )-(-19.304\,(m)/(s) )]

Imp = 11.666\,(kg\cdot m)/(s)

The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?



a. 409.5 m/s b. f₁  = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz


a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5

f₃ = 5v/4L

= 5 × 409.5 m/s ÷ (4 × 0.75 m)

= 5 × 409,5 m/s ÷ 3 m

= 682.5 Hz