For each, find the radius, diameter, and circumference of the circular object (one of these measurements is given in the problem). When working with the circumference, use π and round to the nearest whole number. a. Breaking a cookie in half creates a straight side 10 cm long.radius: cmdiameter: cmcircumference: cm

The type of wave that is a combination of longitudinal and Transverse wave is called Surface wave

The new length of the steel bar is 11.65 cm.

Linear expansion can be defined as the increase in length of a material due to increase in the temperature of the material.

Mathematically, the linear expansion of a material is given as;

ΔL = L₀αΔT

where;

The change in the length of the steel is calculated as follows;

ΔL = (11.5 cm) x (11 x 10⁻⁶/⁰C) x (1221 ⁰C - 22 ⁰C)

ΔL = 0.152 cm

The new length of the steel bar is calculated as follows;

L = ΔL + L₀

L = (0.152 cm) + (11.5 cm)

L = 11.65 cm

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a ) The angle between the ropes when the crate is on the ground = 18.4°

b ) Force exerted by Sean and Greg = 260.8 N

c ) Force exerted when the crate is 5.75 ft below them = 354.9 N

a ) The angle between the ropes,

Distance between crate and a person = 33.5 ft

Distance between Sean and Greg = 11 ft

Consider only one person side of the rope and the midway point between Sean and Greg. This forms a right angled triangle.

sin θ = 5.5 / 33.5

θ = [tex]sin^{-1}[/tex] ( 0.16 )

θ = 9.2°

Angle between the ropes = 2 θ

Angle between the ropes = 2 * 9.2

Angle between the ropes = 18.4°

b ) Force exerted,

Since both ropes pull the same amount of weight for the same amount of distance, their tensions will be equal,

T = T1 = T2

Resolving the tension into its horizontal and vertical component.

[tex]T_{y}[/tex] = T cos θ

[tex]T_{x}[/tex] = T sin θ

Since there is not time component mentioned assuming the crate is pulled at a constant velocity. Therefore acceleration will be zero and hence net force in vertical direction will be zero.

m = 115 lb = 52.16 kg

∑ [tex]F_{y}[/tex] = 0

T cos θ + T cos θ - m g = 0

2 T cos 9.2° = 52.16 * 9.8

T = 511.17 / 1.96

T = 260.8 N

c ) When the crate is 5.75 ft below them,

tan θ = 5.5 / 5.75

θ = [tex]tan^{-1}[/tex] ( 0.96 )

θ = 43.8°

∑ [tex]F_{y}[/tex] = 0

T cos θ + T cos θ - m g = 0

2 T cos 43.8° = 52.16 * 9.8

T = 511.17 / 1.44

T = 354.9 N

d ) The force to lift the crate changes as it moves closer to them is because in a shorter cable the horizontal force increases and the vertical force decreases. So it becomes harder to pull as it gets closer to the destination.

Therefore,

a ) The angle between the ropes when the crate is on the ground = 18.4°

b ) Force exerted by Sean and Greg on the ropes when lifting the 115-lb crate off the ground = 260.8 N

c ) Force exerted by Sean and Greg when the crate is 5.75 ft below them = 354.9 N

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According to Newton's Third Law of Motion, the force that object A exerts to object B has the same magnitude as the force that object B to object A, but in the opposite direction:

Then, the force exerted by Earth on the Moon has the same magnitude as the force exerted by the Moon on the Earth.

Therefore, the given statement is false.

Answer:

a = 1.1 m/s²

Explanation:

Given:

P = 650 N

m = 75 kg

g = 9.8 m/s²

____________

a - ?

Forse:

P = m*(g - a)

650 = 75*(10 - a)

10 - a = 650 /75

Acceleration:

a = 9.8 - 650 / 75 = 9.8 - 8.7 = 1.1 m/c²

Given:

Time taken for 5 rotations = 5.15 seconds

Time for 1 rotation = 1.07 seconds

Distance from shoulder to elbow = 29 cm

Distance from shoulder to the middle of the hand = 57 cm

Let's use the information above to answer the following questions.

Question 2:

Let's determine how far in degrees the hand travelled during the five rotations.

In one full rotation, we have 360 degrees.

Thus, 5 full rotations = 5 * 360 = 1800 degrees

Therefore, in 5 full rotations, the hand travelled 1800 degrees.

B. In radians, we have:

180 degrees = π rad

C. To find the distance in meters, we have:

Distance from elbow to shoulder = 29 cm = 0.29 meters

Therefore, the hand travelled 9.11 meters during the five rotations.

Question 3:

A. To find the average angular speed, apply the formula:

Where t = 5.15 seconds

Thus, we have:

B. Average linear speed of the hand.

To find the average linear speed of the hand, we have:

C. The average angular speed and average linear speed are the same

time = 27 s

d = 20 yard

a) speed = distance / time

d1 = 100 yard ( own goal to opposing team's goal line)

d2 = 80 yard ( returning to 20 yard line)

d= d1+d2

d= 100 + 80 = 180 yards

Speed = 180 y / 27s = 6.667 y/s

b) velocity = displacement / time

d = 100 - 80 = 20 y

Velocity = 20 / 27 = 0.74 y/s

The cat’s speed when it slid off the table will be 6.552 m/s

The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.

a = -g = 9.8 m[tex]/s^{2}[/tex]

using equation of motion

x = u(horizontal )*t + 1/2 * a (horizontal) * [tex]t^{2}[/tex]

since , a (horizontal) = 0

x = u(horizontal )*t

u   = x / t                   equation 1

similarly

y = u(vertical)*t + 1/2 * a (vertical) * [tex]t^{2}[/tex]

u(vertical) = 0

t = [tex]\sqrt{2y / a}[/tex]                               equation 2

substituting the value of equation 2  in equation 1  

u = x /  [tex]\sqrt{2y / a}[/tex]  

= [tex]\sqrt{\frac{-9.81}{2*-0.66} } * 2.4[/tex]

= 6.552 m/s

The cat’s speed when it slid off the table will be 6.552 m/s

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Note that, an isolated system does not allow:

• the exchange of energy

• the exchange of matter

Therefore, in an isolated system, neither

hea

Answer:

Explanation:

c. Alone

Answer:

C. alone.

Explanation:

The velocity of a book that fell off a table and hit the ground after 5 seconds is 49 m/s.

Velocity can be defined as the ratio of displacement to time.

To calculate how fast the book will hit the ground, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the velcoity of the book before it hit the ground is 49 m/s.

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Answer:

Explanation:

a) 301 cg

b) 6.2 km

c) 0.00013

Given:

The mass of thee person is

The height at which person standing is

Required: calculate the potential energy of the person

Explanation:

when anybody of mass m is at a distance h from the earth's surface then it has potential energy that is given as

where g is the acceleration due to gravity whose value is

Plugging all the values in the above relation, we get;

Thus, the potential energy is

Use the radioactive decay formula using t_0 as the half life:

Substitute the initial amount of radioactive element A_0=100g, the half life t_0=10y and the time period t=20y to find the remaining amount after 20 years:

Therefore, after 20 years, there are 25 grams left.

Given data:

* The mass of the Bart is m_1 = 32.4 kg.

* The mass of the Milhouse is m_2 = 27.6 kg.

* The distance between the Millhouse and Bart is d = 4 m.

Solution:

To balance the seesaw, the net moment about the center should be zero.

The diagrammatic representation of the given system is,

The distance between the Bart and Milhouse can be written as,

where d_2 is the distance of Milhouse from the center and d_1 is the distance of Bart from the center,

Consider the moment as positive if it is in an anticlockwise direction and negative if it is a clockwise direction.

Thus, the net moment about the center is,

Substituting the known values,

By simplifying,

Thus, the distance of the Bart from the center is 1.84 meters.

We will use Snell's law, which states:

Where n1 and n2 are the refraction indexes and their respective angles are "theta1" and "theta2".

For part A we replace the values:

Now we solve for "theta2" first by dividing both sides by 1.33:

Now we use the inverse function for sine:

Solving the operation:

For part B, since "theta1" and "theta3" are angles of reflection, according to the reflection law, these angles are equal, therefore:

For part C. The index of refraction is defined as:

Where "c" is the speed of light in a vacuum and "v" is the speed of light in the medium. Replacing the values:

Now we solve for "v":

Solving the operation:

For part d. We will use the following formula:

Where "v" is the speed and "f" is the frequency. Replacing we get:

Solving the operations:

For part e. The largest value of the angle of incidence that will result in refraction is 90 degrees.

Answer:

20 mL

Explanation:

The volume of the rock is equal to the difference of volume of A and B. So, it is equal to

90 mL - 70 mL = 20 mL

Because 90 mL is the volue in cylinder B and 70 mL is the volume in cylinder A.

Therefore, the volume of the rock is 20 mL

a. Pulley I

Explanation

A pulley system is a collection of one or more wheels which are used with a. rope or chain to make it easier to lift things, The main advantage in the use of pulleys is that the effort becomes less as compared to the normal lifting of the weights

the ideal mechanical advantage is equal to the number of rope segments pulling up on the object. The more rope segments that are helping to do the lifting work, the less force that is needed for the job.

so

Step 1

let's check the number of pulleys on each system

so, the firs system has the more rope segements pulling up (4)

therefore, the answer is

a. Pulley I

I hope this helps you

The linear velocity of the ball whose angular velocity is 31.3 rad/s about the elbow joint will be 45.385 m/s.

Angular velocity is the rate of change of angular displacement with respect to time. Mathematically -

ω = dθ/dt

from this we can write -

dθ = ω dt

∫dθ = ω ∫dt

θ₂ - θ₁ = ω(t₂ - t₁)

Δθ = ω Δt

ω = Δθ/Δt

Given is a ball thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. The angular velocity of the ball about the elbow joint is 31.3 rad/s and the ball is 1.45 m from the elbow joint.

The relation between linear velocity and the angular velocity of a body undergoing circular motion is given by -

v = rω

From this we can write -

ω = 31.3 rad/s

r = 1.45 m

Substituting the values -

v = rω

v = 1.45 x 31.3

v = 45.385 m/s

Therefore, the linear velocity of the ball whose angular velocity is 31.3 rad/s about the elbow joint will be 45.385 m/s.

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The maximum mass of the barrel, with maximum torque of 3.0×10° N.m delivered about the axis is 0.102 kg.

Mass is the quantity of matter a body contains. The S.I unit of mass is kilogram (kg). Mass can also be defined as the ratio of the force to the acceleration of a body.

To calculate the maximum mass of the barrel, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute the values above into equation 1

Hence, the  maximum mass of the barrel is 0.102 kg.

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A 2000 kg car is stopped by applying a braking force of 5000 newtons. The acceleration caused by this braking force is (a)=2.5 m/s²

When a object start with a velocity and ends with different velocity, so the change in velocity in a given time is called the acceleration of the object. It is a vector quantity. It can be measured in m/s².

To calculate the acceleration we are using the formula here is ,

F=ma

Here we are given by the question is,

F= The amount of force applied on the car. = 5000N.

m= The mass of the car. = 2000kg

We have to calculate the change in acceleration = a m/s².

Now we put the values in above equation we get,

F=ma

Or, a= F/m

Or, a=5000/2000

Or, a=2.5 m/s².

Now from the above calculation we can say that, The acceleration caused by this braking force is (a)=2.5 m/s².

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The perfectly straight lines of reflection demonstrated in the nature is due to particle nature of the light. Therefore, Newton described that the light consists of 'particles' which means second option is correct.

We are given that two objects are being attracted by a gravitational force between each other of 10N. The gravitational force between two masses is given by the following equation:

Where:

Replacing the given values for the 10N force:

Where:

Now we will solve for the product of the masses and the gravitational constant by multiplying both sides by the distance squared:

Now, the product of the masses and the gravitational constant won't change if we double the distance, therefore, if we apply the equation for the gravitational force for the new distance we get:

We can replace the value we determined earlier:

Since the distance is double, we have:

Replacing in the previous equation:

Solving the square:

Now we can cancel out the distances squared:

Solving the operation:

Therefore, doubling the distance the new gravitational force is 2.5N.

The change that would increase the amount of mass is Increase in the number of bricks resting upon the cart. That is option A.

Mass is defined as the amount or quantity of matter that makes up an object which can be measured in grams and Kilograms.

The mass of an object is dependent on the change of its weight, that is, an increase in the weight of an object increases its mass.

Since the bricks had the same mass as the cart, adding one brick to the cart would double the mass of the object. Adding two bricks to the cart would triple the mass of the object.

That is to say, when more bricks are being laid on the cart, it's weight increase leading to an increase in its mass also.

Therefore, it can be concluded that increase the amount of mass is Increase the number of bricks resting upon the cart.

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ANSWER:

39.1 km

STEP-BY-STEP EXPLANATION:

To better understand the problem, we make a sketch, like this:

Therefore, we can determine the distance east with the cosine function, like this:

The distance to the east is 39.1 km

a.) the acceleration when the car is at B located at an angle of 35.0° is 2.689i -0.42818j

b.) the car's average speed is  v= 6.84375 m/s

c.) average acceleration during the 32.0-s interval is (−0.181 i+0.181 j)m/s²

Acceleration is described as the rate of change of the velocity of an object with respect to time.

(a) The car’s speed around the curve is found from

v= 219/32.0

v= 6.84375 m/s

This is the answer to part (b) of this problem. We calculate the radius of the curve from

(1/4) X 2πr = 219 m

​ which gives r = 139.4 m

The car’s acceleration at point B is then

ar =   (V²/r ) towards the center

=  ( 6.84375)² / 139.4 at 35.0° north of west

=  (2.9761 m/s²)(cos 35.0)(-i) + (sin 35.0j)

= -2.689i -0.42818j

(b) From part (a), v= 6.84375 m/s

(c) We find the average acceleration from

A avg = (Vf - Vi)/ change in t

A avg = ( 6.84375 j -  6.84375 i ) / 32.0 s

A avg = (−0.181 i+0.181 j)m/s²

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Given data

*The given initial velocity of the puck is u = 5 m/s

*The given final velocity of the puck is v = 30 m/s

*The given distance is s = 10 m

The formula for the acceleration is given by the kinematic equation of motion as

Substitute the known values in the above expression as

Hence, the acceleration of the puck is a = 43.75 m/s^2

We know that the amount of matter is given by:

where λ is the decay constant. The decay constant is related to the half-life of the element by the equation:

Then we can express our first equation as:

Plugging the initial amount, 50 g, the half-life of 5715 years and the time we want to know we have that:

Therefore, after 1600 years there are 41.181 g

Explanation

A longitudinalwave is in which the particles of the medium vibrate in the direction of the line of advance of the wave.Longitudinal waves cause the medium to move parallel to the direction of the wave.

A longitudinal wave can be set up for example in a streched spring by compressing the coils in a small region, and releasing the compressed region,

the back and forth motions of the coils of the spring is in the same direction that the wave travels

so, in a longitudinal wave there is not Vertical movement, so the answer is

C. Vertical

Considering the given position function, it is found that:

a) The velocity function is: v(t) = 5 m/s.

b) The functions are graphed at the end of the answer.

The position function in this problem, after t seconds, is defined according to the following rule:

s(t) = 5t.

The velocity function is the derivative of the position function, hence it is calculated as follows:

v(t) = s'(t) = [5t]' = 5 m/s. (position in meters, hence velocity in meters per second).

The derivative rule applied was the power rule, [5t]' = 5[t'] = 5 x 1 x t^(1 - 1) = 5t^0 = 5.

These two functions are graphed at the end of the answer, considering a domain of t ≥ 0, as time cannot assume negative values.

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