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4 years ago

12

When the height of the cylinder increased by a factor of 5, from 100 m to 500 m, what happened to the cylinder’s gravitational p

otential energy?

2 answers:

Grace [21]4 years ago

8 0

PE = m*g*h ⇒ 5* PE = 5*m*g*h

storchak [24]4 years ago

4 0

It increased by a factor of 5

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Determine the displacement and distance covered by a man if he walks 10 m north, turns east and walks 20 m, and then turns right

NemiM [27]

Answer:

20m

Explanation:

The two tens cancel each other out, as they are in opposite directions. Now we only care about the 20m, which if we have no 10's, will end up 20m away.

7 0

3 years ago

7. When will an object's displacement and distance traveled be different?

mariarad [96]

If an object changes direction while travelling will an object's displacement and distance travelled be different.

Some people believe that distance and displacement are simply different names for the same quantity. However, distance and displacement are not the same thing. If an object changes direction while travelling, the total distance travelled is greater than the displacement between those two points.

The magnitude of the displacement is always less than or equal to the distance because it is measured along the shortest path between two points.

When the direction of displacement does not change, the magnitude of the displacement and distance are the same. When a body travels in a straight line, for example, its displacement and distance are the same.

Learn more about displacement and distance brainly.com/question/3243551

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8 0

2 years ago

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

6 0

3 years ago

The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. For the same object in low-earth orbit arou

mojhsa [17]

Answer:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

$g = \dfrac{GM}{r^2}$

$g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}$

g = 8.61 m/s²

3 0

3 years ago

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

VMariaS [17]

Answer:

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 \times 10^{-2} N$

$Q = 6.80 \times 10^3 C$

now we have

$6.57 \times 10^{-2} = (6.80 \times 10^3)E$

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

4 0

4 years ago