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Sam is measuring the velocity of a car at different times. After two hours, the velocity of the car was 50km/h. After six hours, the velocity of the car is 54km/h. Part A: Write an equation in two variables in standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used. (5points) Part B: How can you graph the obtained in Part A for the first seven hours? (5points)

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Answered by answersmine AT 22/10/2019 – 02:54 AM

You are given the time and speed so from there you can get the points (2,50) and (6,54) once you got that you can see how the slope is 1 and 
b=48
x-y=-48 (standard form)
 
part B= 
1    49
2    50
3    51
4    52
5    53
6    54
7    55

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Use the Pythagorean theorem to answer this question. Becca paddles a boat from the south bank of a stream to the north bank. She paddles at a rate of 8 mph. The stream is flowing west at a rate of 6 mph. What is Becca’s actual velocity?

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Since a stream of water is perpendicular to the movement
of Becca, her boat must have been moving going at an angle of north east. Using
the Pythagorean formula, we can calculate for the resultant velocity, c.

c^2 = (8 mph)^2 + (6 mph)^2

c^2 = 64 + 36

c^2 = 100

c =
10 mph

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A golf ball is hit from the ground with an initial velocity of 208 ft per second assume the starting height of the ball is 0 ft how long will it take the golf ball to hit the ground

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A golf ball is hit from the ground with an initial velocity of 208 ft per second assume the starting height of the ball is 0 ft how long will it take the golf ball to hit the ground

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Bret kicks a soccer ball straight up into the air worth an initial velocity of 25 m/s from a height of 1 m. After how many seconds will the soccer ball land? A. 5 s B. 5.14 s C. 6.5 s D. 7 s

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Bret kicks a soccer ball straight up into the air worth an initial velocity of 25 m/s from a height of 1 m. After how many seconds will the soccer ball land? A. 5 s B. 5.14 s C. 6.5 s D. 7 s

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If a wheel with a radius of 80 inches spins at a rate of 50 revolutions per minute, find the approximate linear velocity in miles per hour

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If a wheel with a radius of 80 inches spins at a rate of 50 revolutions per minute, find the approximate linear velocity in miles per hour

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If a wheel with a radius of 30 inches spins at a rate of 500 revolutions per minute, find the approximate linear velocity in miles per hour.

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30 inches @ 500 rpm = 15,000 inches per minute

15,000 inches per minute = 900,000 inches per hour (multiply by 60)

1 mile = 5,280 feet (google-able fact you shouldn’t have to memorize)
1 foot = 12 inches (you probably already know this)
900,000 inches = 75,000 feet (divide by 12)

75,000 feet = 14.20454545454545. . . miles (divide by 5280)

Linear velocity in miles per hour is 14.20454545454545. . . miles per hour.

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Which statement best describes how work and power are different? A)To find work we need to know force and distance; to find power we need to know force and velocity. B) To find work we need to know energy and time; to find power we need to know energy and distance. C)To find work we need to know velocity and distance; to find power we need to know distance and time. D)To find work we need to know distance and force; to find power we need to know energy and velocity.

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Which statement best describes how work and power are different? A)To find work we need to know force and distance; to find power we need to know force and velocity. B) To find work we need to know energy and time; to find power we need to know energy and distance. C)To find work we need to know velocity and distance; to find power we need to know distance and time. D)To find work we need to know distance and force; to find power we need to know energy and velocity.

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What is the velocity of an object dropped from a height of 180 m when it hits the ground?

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What is the velocity of an object dropped from a height of 180 m when it hits the ground?

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A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. what is its angular velocity (in rev/s) after a 22.0 kg child gets onto it by grabbing its outer edge? the child is initially at rest.

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To answer this problem, we must remember that momentum is
conserved. Therefore,

Initial momentum = Final momentum

In this case we use angular momentum which it is defined
as:

Momentum =0.5 m ω^2

So finding for final angular velocity, ωf:

mi * ωi^2
= mf * ωf^2

120 kg * (0.5 rev / s)^2 = (120 kg + 22 kg) * ωf^2

ωf^2
= 0.2113

ωf = 0.46 rev/s

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An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it’s going to land. Which term describes the slowing of the plane? 1.stationary positive velocity 2.positive acceleration 3.negative acceleration 4.constant velocity

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The equation of the car is given by the equation,

                          x(t) = 2.31 + 4.90t² – 0.10t⁶

If we are going to differentiate the equation in terms of x, we get the value for velocity.

                  dx/dt = 9.8t – 0.6t⁵

Calculate for the value of t when dx/dt = 0.

                 dx/dt = 0 = (9.8 – 0.6t⁴)(t)

The values of t from the equation is approximately equal to 0 and 2. 

If we substitute these values to the equation for displacement,

(0)   , x = 2.31 + 4.90(0²) – 0.1(0⁶) = 2.31

(2)    , x = 2.31 + 4.90(2²) – 0.1(2⁶) = 15.51

Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters. 

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The police department is very proud of their new police cars. They announced that they are able to go from a stopped position to 80 miles per hour in 6 seconds. They are describing their cars’ velocity. direction. speed. acceleration.

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The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. 

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. 

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h. 

So the time for the pencil and eraser to touch is T=1.3449 hours. 

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

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If a wheel with a radius of 70 inches spins at a rate of 600 revolutions per minute, find the approximate linear velocity in miles per hour.

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If a wheel with a radius of 70 inches spins at a rate of 600 revolutions per minute, find the approximate linear velocity in miles per hour.

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Block a, with a mass of 2.0kg, moves along the x axis with a velocity of 5.0m/s in the positive x direction. it su?ers an elastic collision with block b, initially at rest, and the blocks leave the collision along the x axis. if b is much more massive than a, the speed of a after the collision is:

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The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. 

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. 

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h. 

So the time for the pencil and eraser to touch is T=1.3449 hours. 

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

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A blue billiard ball with mass m crashes into a red billiard ball with the same mass that is at rest. The collision results in the blue billiard ball traveling with a velocity of v and the red billiard ball travelling with a velocity of 3v. In terms of v and m, determine the blue billiard ball’s velocity before the collision…

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Ahemmm having x-intercepts of -3 and -5.. ..well, that simply means, -3 and -5 are roots or solutions or zeros of the equation

it namely means x = -3 and x = -5, an x-intercept is when the graph touches the x-axis, at that point, the y-intercept is 0, so the point is (-3, 0) and (-5, 0)

if the roots are -3, and -5, then

bf begin{cases}
x=-3implies x+3=0implies &(x+3)=0\
x=-5implies x+5=0implies &(x+5)=0
end{cases}\\
-------------------------------\\
(x+3)(x+5)=0implies (x+3)(x+5)=yimplies x^2+8x+15=y

if you have the zeros/x-intercepts/solutions of the polynomial, all you have to do is, get the factors, as above, and get their product, to get the parent original polynomial.

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A 227 kg motorcycle moves with a velocity of 8 m/s. What is its kinetic energy? _______ Joules

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The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. 

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. 

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h. 

So the time for the pencil and eraser to touch is T=1.3449 hours. 

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

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A worker does 500 J of work on a 10 kg box.If the box transfers 375 J of heat to the floor through the friction between the box and the floor,what is the velocity of the box after the work has been done on it?

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The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. 

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. 

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h. 

So the time for the pencil and eraser to touch is T=1.3449 hours. 

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

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A woman rides a bicycle for 3 hours and travels 51 kilometers. Find the angular velocity of the wheel if the radius is 30 centimeters

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A woman rides a bicycle for 3 hours and travels 51 kilometers. Find the angular velocity of the wheel if the radius is 30 centimeters

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Determine how long it will take an object to rotate through 10.0 revolutions at an angular velocity of 4.8 radians per second. use . round the answer to the nearest tenth.

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Answer:

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

Explanation:

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t²      (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²    

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the  equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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Determine the linear velocity of an object with an angular velocity of 5.9 radians per second at a distance of 12 centimeters from the center. use . round the answer to the nearest tenth.

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Answer:

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

Explanation:

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t²      (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²    

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the  equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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An object with an initial velocity of 3.0 m/s has a constant acceleration of 2.0 m/s2. When its speed is 19.0 m/s, how far has it traveled?

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The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. 

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. 

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h. 

So the time for the pencil and eraser to touch is T=1.3449 hours. 

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

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A bowling ball weighs 7.26 kg and takes 3.64 seconds to travel the 19.2 m length of the bowling alley. assuming the velocity is constant, what is the kinetic energy of the bowling ball?

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The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. 

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. 

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h. 

So the time for the pencil and eraser to touch is T=1.3449 hours. 

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

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A railroad freight car with a mass of 29,000 kg is moving at 2.2 m/s when it runs into another car with a mass of 30,000 kg the second car is at rest the cars locked together what is the final velocity

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The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. 

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. 

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h. 

So the time for the pencil and eraser to touch is T=1.3449 hours. 

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

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Rebekah kicks a soccer ball off the ground and in the air, with an initial velocity of 25 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches? 9.5 feet 9.8 feet 10.2 feet 10.7 feet

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Rebekah kicks a soccer ball off the ground and in the air, with an initial velocity of 25 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches? 9.5 feet 9.8 feet 10.2 feet 10.7 feet

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. Which of the following statements is accurate? A. If an object’s velocity is changing, it’s either experiencing acceleration or deceleration. B. If an object’s velocity decreases, then the object is accelerating. C. If an object is said to be decelerating, its velocity must be increasing. D. If an object’s velocity remains constant, its acceleration must be increasing.

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Answer:

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

Explanation:

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t²      (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²    

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the  equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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A spinning spherical shell with radius 30 cm and negligent mass is filled with 0.5kg of sand. It started spinning at an angular velocity of 8 m/s. All the sand starts within the center 10-cm and by the end has moved to the outermost radius. What is its new angular velocity?

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The gravitational force Fg between two objects is given by the equation:

Fg=(G*m₁*m₂)/r₂, where G=6.67*10^-11 m³ kg⁻¹ s⁻² is the gravitational constant, m₁ and m₂ are the masses of the two bodies and r is the distance between those bodies. 

Due to the gravitational attraction the pencil and the eraser will attract if we there is no friction on the surface. 

m₁=10 g=0.01 kg is the mass of the pencil
m₂=20 g=0.02 kg is the mass of the eraser
r=2.5 cm = 0.025 m

First we calculate the Fg:

Fg={(6.67*10^-11)*0.01*0.02}/(0.025²)=2.1344*10^-11 N

To get the velocity v of the pencil:

v²=2as, where a is the acceleration of the pencil and s is the path. In our case s=r so we can write:

v²=2ar

a=Fg/m₁= 2.133*10^-9 m/s²

v²=2*(2.133*10^-9)*0.025=1.0665*10^-10

v=√(1.0665*10^-10)=1.0327*10^-5 m/s

We have the velocity and the acceleration, so we can calculate the time t with the equation:

t=v/a=(1.0327*10^-5)/(2.133*10^-9)=4841.6 s

1 hour has 3600 s so when we divide time t in seconds by 3600 we get time T in hours:

T=t/3600=4841.6/3600=1.3449 h. 

So the time for the pencil and eraser to touch is T=1.3449 hours. 

Also time T can be expressed like T= 1h and 20 mins and 41.64 s

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