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3 years ago

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

Physics

1 answer:

Lesechka [4]3 years ago

6 0

Answer: It increased by a factor of 5

Explanation:

Gravitational potential energy $P$ is given by the following formula:

$P=mgh$

Where:

$m$ is the mass of the cylinder (assumed as constant)

$g$ is the acceleration due gravity (assumed as constant)

$h$ is the height of the cylinder

As we can see, $P$ is directly proportional to $h$ . Hence, if $h$ increases by a factor of 5 $P$ increases by a factor of 5, as well.

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An object with an acceleration of –10 m/s² is _____.

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3 years ago

Wo siblings are arguing over what way to pull a 10 kg wagon. Boris wants to pull it to the right. Boris puts 220 N of force on t

Mrrafil [7]

Answer:

a) Then net force acting on the wagon 30 [N] in a negative direction to the left

b)Acceleration of the wagon 3 m/s² in a negative direction to the left

Explanation:

The coordinates x and y show the positive direction in x-axis ( to the right) and negative direction to the left. In north-south direction positive y values are to the north and negative values to the south

The free-body diagram shows 4 forces acting

In y-axis:

The weight ( P = m*g = 10 [kg]*9,8 [m/s²] = 98 [N] negative

Normal reaction force Fn = P = 98 [N] positive ( surface is frictionless)

∑Fy = 0 P - Fn = 0 P = Fn = 98 [N]

In x-axis:

Boris pulling force to the right (positive ) 220 [N]

Natasha pulling force to the left ( negative) 250 [N]

∑Fx = m*a

220 [N] - 250 [N] = 10 Kg * a [N] = Kg * m /s²

-30 [kg*m/s²] = 10 * Kg * a

-30/10 = - 3 m/s² = a

Sign (-) means the direction of acceleration vector is to the left (the same direction of the movement )

Then net force acting on the wagon 30 [N] in negative direction to the left

Acceleration of the wagon 3 m/s² in negative direction to the left

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3 years ago

Why do scientists believe that dark matter exists even though it cannot be seen?

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3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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3 years ago