Answer:

**Answer:**

v = 0.85 m/s

**Explanation:**

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 = 0.85 m/s**

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

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy of the ball and the air in the system increases by 15 J. Just before it hits the surface its speed is

Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Part A A microphone is located on the line connecting two speakers hat are 0 932 m apart and oscillating in phase. The microphone is 2 83 m from the midpoint of the two speakers What are the lowest two trequencies that produce an interflerence maximum at the microphone's location? Enter your answers numerically separated by a comma

Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.

A large rectangular tub is filled to a depth of 2.60 m with olive oil, which has density 915 kg/m3 . If the tub has length 5.00 m and width 3.00 m, calculate (a) the weight of the olive oil, (b) the force of air pressure on the sur- face of the oil, and (c) the pressure exerted upward by the bottom of the tub.

Do you think a baseball curves better at the top of a high mountain or down on a flat plain

Part A A microphone is located on the line connecting two speakers hat are 0 932 m apart and oscillating in phase. The microphone is 2 83 m from the midpoint of the two speakers What are the lowest two trequencies that produce an interflerence maximum at the microphone's location? Enter your answers numerically separated by a comma

Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called a. open pneumothorax. b. empyema. c. pleural effusion. d. tension pneumothorax.

A large rectangular tub is filled to a depth of 2.60 m with olive oil, which has density 915 kg/m3 . If the tub has length 5.00 m and width 3.00 m, calculate (a) the weight of the olive oil, (b) the force of air pressure on the sur- face of the oil, and (c) the pressure exerted upward by the bottom of the tub.

Answer:

Time = 1.61 seconds

Explanation:

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

**Answer:**

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

**Explanation:**

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

using

-----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

= 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**

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.**

#SPJ12

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.

#SPJ11

**Answer:**

s = 6.25 10⁻²² m

**Explanation:**

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.

**Given Information:**

Mass of ball = m = 0.320 kg

Initial height = h₁ = 19 m

Final height = h₂ = 15 m

**Required Information:**

Impulse = I = ?

**Answer:**

Impulse = 11.77 kg.m/s

**Explanation:l**

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

**Answer:**

**Explanation:**

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

**Before collision**:

**After collision**:

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

**Answer:**

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

**Explanation:**

**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