# 2. A book weighs 6 newtons. A librarian picks up the book from one shelf and puts it on a shelf 2 metres higher. (a) Calculate the work done on the book. [Show your working]. (b) The next person to take the book from the shelf accidentally drops it. The book accelerates at 9.8m/s². Use this information to calculate the mass of the book. [Show your working].

3 weeks ago

## Solution 1

Guest #7253028
3 weeks ago

0.612 kilograms.

Explanation:

To solve this problem, we need to use the definition of work as the product of force and displacement.

(a) The work done on the book is the force applied to it multiplied by the distance through which it is moved. In this case, the force applied to the book is its weight, which is 6 newtons. The displacement of the book is 2 meters, because it is moved from one shelf to another shelf that is 2 meters higher. Therefore, the work done on the book is 6 newtons * 2 meters = 12 joules.

(b) The next person to take the book from the shelf accidentally drops it. When an object is dropped, it accelerates due to the force of gravity, which is 9.8 m/s². Since the force of gravity is the same for all objects, we can use the acceleration of the book to calculate its mass. Specifically, we can use the formula F = ma, where F is the force acting on the object (in this case, the force of gravity), m is the mass of the object, and a is its acceleration.

In this case, the force acting on the book is its weight, which is 6 newtons. The acceleration of the book is 9.8 m/s². Therefore, we can solve for the mass of the book using the formula F = ma: 6 newtons = m * 9.8 m/s². This simplifies to m = 6 newtons / 9.8 m/s² = 0.612 kilograms.

In summary, the work done on the book when it is picked up and moved to a higher shelf is 12 joules, and the mass of the book is 0.612 kilograms.

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How does the restoring force and the mass of the string effect the speed of a wave generated on a string?
Solution 1

The string will extend when the object is moved to the right, applying a restoring force to bring the weight back to its equilibrium position. The string will compress if the object is pushed to the left, and it will then exert a restoring force to bring the thing back to its starting position.

### What is the impact of mass on restoring force?

The displacement diminishes as the mass comes closer to the equilibrium position, and as a result, the restoring force and acceleration both decrease. There is no restoring force once the mass has reached equilibrium.

### What influences how quickly waves go down a stretched string?

The linear density of the string and the tension in the string both affect how quickly a wave moves along a string. A string's linear density is its mass per unit length. In general, a wave's speed is determined by the square root of the medium's elastic property to inertial property ratio.

### How does string mass affect the wave's speed?

The speed would be higher in the low linear mass density of the string because the speed of a wave on a string is inversely proportional to the square root of the linear mass density.

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Question
A 500 kg empty elevator moves with a downwards acceleration of 2 m/s2. What is the tension force of the cable on the elevator?
Solution 1

The tension force of the cable on the elevator is 4000N.

From the question,

Mass of an elevator(m) = 500kg

It is accelerating downhill at a rate of $$2\frac{m}{s^{2} }$$. It is necessary to determine the tension force of the cable on the elevator. We know that the net force on the elevator when it is traveling downhill is provided by:

F = m(g - a)

Here, g is the acceleration due to gravity

g = $$9.8\frac{m}{s^{2} }$$

g ≅ 10$$\frac{m}{s^{2} }$$

F = 500 × (10 - 2)

F = 500 × 8

F = 4000N

Therefore, the tension force of the cable on the elevator is 4000N.

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4. A 20-kg box sits on an incline of 30° from the horizontal. The coefficient of friction between the box and the incline is o.30. Find the acceleration of the box down the incline.
Solution 1

The box accelerates down the incline at 2.4m/s^2.

### How do you find the acceleration of an incline box?

A mass m particle will slide down a smooth inclined plane if it is released onto it with a friction force F=0. We resolve in the direction of motion to determine the particle's acceleration as it slides. F = ma, mg cos(90) = ma, gg cos(90) = a, gg sin() = a.

By using zigmaFx = max, we will determine the acceleration.

However, we must first determine the friction force Ff.

Because cos 30 degree = 0.866 and Fy = ma y = 0 result in FN - 0.87mg = 0, we may calculate FN as (0.87)(20Kg)(9.81 m/s2) = 171N.

From Ff = mue FN = 0.30)(171 N)= 51N, we can now calculate Ff.

We get Ff - 0.50mg = ma x 51N - (0.50)(20)(9.81)N = (20kg)(ax) from the expression zigmaFx = max, where ax = -2.35 m/s2.

At 2.4 m/s2, the box quickens its descent of the hill.

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Question
You are a crash test analyst for Toyota. Car 1 (mass = 900 kg) is at rest while Car 2 (mass = 1700 kg) is traveling toward Car 1 at 15.0 m/s. The two cars collide and and their bumpers stick together. What is the resulting momentum after the collision?
Solution 1

The momentum after the collision is 25500 kg⋅m/s. Each object's momentum may change, but the overall momentum must not change.

### What is Momentum ?

In Newtonian mechanics, momentum is calculated as the sum of an object's mass and velocity. It has a magnitude and a direction, making it a vector quantity.

The rate of change of a body's momentum is equal to the net force exerted on it, according to Newton's second law of motion. Regardless of the frame of reference, momentum is a conserved quantity in any inertial frame, which means that if an enclosed system is not subject to outside influences, its total linear momentum remains constant.

The formula for calculating momentum is expressed as

momentum = mass x velocity

According to the law of conservation of momentum, initial momentum = final momentum

The formula for initial momentum is

m1u1 + m2u2

where

m1 = mass of first car

u1 = initial velocity of first car

m2 = mass of second car

u2 = initial velocity of second car

From the information given,

m1 = 900

u1 = 0 because it is at rest

m2 =  1700

u2 = 15.0 m/s

Initial momentum = 900 x 0 +  1700 x 15.0 = 25500

Momentum after the collision is 25500 kg⋅m/s

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I don't see a picture attached to your question. Without one, I don't think I will be able to determine what type of orbit is displayed.

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On a train moving , 75 m/s you roll a ball towards the front of the train with a speed of 5m/s. Your friend standing some unknown distance away from you catches it. To a person outside the train, the ball traveled 220m until it was caught. How far away is your friend standing from you?​
Solution 1

The distance of your friend from you is 15.7 m.

### What is the time of motion of ball?

The time take for the ball to travel to you friend is calculated by applying the principle of relative velocity as shown below.

( Vt - Vb ) t = d

where;

• Vt is the velocity of the train
• Vb is the velocity of the ball
• t is the time of motion of the ball
• d is the distance travelled by the ball

( Vt - Vb ) t = d

(75 - 5) t = 220

70t = 220

t = 220 / 70

t = 3.14 s

The distance of your friend from you is calculated as;

d1 = 5 m/s x 3.14 s

d1 = 15.7 m

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Question
A star was observed to have a planet orbiting it at a distance of 1.12×108 km. The orbital period of the planet is 329 days. Calculate the mass of the star.
Solution 1

The mass of the star would be 9.97 x 10³⁵ Kg.

### What is Star? What is centripetal force?

• Stars are huge celestial bodies made mostly of hydrogen and helium that produce light and heat from the churning nuclear forges inside their cores.
• We can write the relation absolute magnitude [M] and apparent magnitude [m] as follows : M = m + 5(log₁₀p + 1).
• The centripetal force is the force needed to make a body move in circular motion. It is given by : F{C} = mv²/r

We have a star was observed to have a planet orbiting it at a distance of 1.12 × 10⁸ km. The orbital period of the planet is 329 days.

Let the mass of star be [M] and that of planet be [m]. Since, the planet revolves around the star, we can write -

F{Centripetal} = F{Gravitation}

mv²/r = GMm/r²

v² = GM/r

(rω)² = GM/r

r²ω² = GM/r

ω² = GM/r³

(2π/T)² = GM/r³

M = (2π/T)² (r³)/G

M = 46.8 x (r³)/G

M = {46.8 x (1.12)³ x (10⁸)³}/{(6.6) x 10 ⁻¹¹}

M = (65.7/6.6) x (10²⁴/ 10 ⁻¹¹)

M = 9.97 x 10³⁵ Kg

Therefore, the mass of the star would be 9.97 x 10³⁵ Kg.

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how was the third law of motion discovered
Solution 1

how was the third law of motion discovered

Explanation:

Isaac Newton discovered this law by exerting pressure on one object against another. The first object's force is felt and equally replicated by the second body. Thus, Isaac Newton proved that it's impossible to exert pressure on a single object without a similar reaction from the second object.

Solution 2
Isaac Newton discovered this law by exerting pressure on one object against another. The first object's force is felt and equally replicated by the second body. Thus, Isaac Newton proved that it's impossible to exert pressure on a single object without a similar reaction from the second object.
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Which statement best describes the relationship between changes in air pressure and wind speeds?

Responses

When a high- and a low-pressure air mass are far apart, air moves quickly from high to low pressure.

When a high- and a low-pressure air mass are far apart, air moves slowly from high to low pressure.

When a high- and a low-pressure air mass are close together, air moves slowly from high to low pressure.

When a high- and a low-pressure air mass are far apart, air moves quickly from low to high pressure.
Solution 1

The best statement that best describes the relationship between changes in air pressure and wind speeds is when a high- and a low-pressure air mass are far apart, air moves quickly from high to low pressure.

The greater the increase or decrease in pressure, the faster the winds move. When a high- and a low-pressure air mass are far apart, air moves slowly from high to low pressure. When a high- and a low-pressure air mass are close, the air moves slowly from a point of high to low pressure. When a high- and a low-pressure air are far apart, the air moves rapidly from a region of high pressure to a region of low pressure.

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