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3 years ago

B. Calculate the rock's gravitational potential energy at 50 m, 20 m, 1 m, and om high. Put the

Physics

1 answer:

klasskru [66]3 years ago

4 0

The gravitational potential energy of the rock is:

490 J at 50 m

196 J at 20 m

9.8 J at 1 m

Explanation:

The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in the gravitational field.

For an object near the Earth's surface, the GPE is given by

$GPE=mgh$

where

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity at Earth's surface

h is the heigth of the object relative to the ground

For the rock in this problem,

m = 1 kg

So its GPE at h = 50 m is

$GPE(50)=(1)(9.8)(50)=490 J$

The GPE at h = 20 m is

$GPE(20)=(1)(9.8)(20)=196 J$

While the GPE at h = 1 m is

$GPE(1)=(1)(9.8)(1)=9.8 J$

Learn more about gravitational potential energy:

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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

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Therefore;

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On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

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a·x² + c·v² = b

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a, b, and c are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.

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The possible question options are;

As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x

The relationship is no longer linear and v will be more for the same value of x

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2 years ago

Arunner has a speed of 23 m/s. They see the finish line and speed up to 27 m/s. This happens in 5 seconds. If the runner has a m

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2 years ago

A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J

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Explanation:

From the given information:

Let the initial kinetic energy from point A be $K_A$ = 1.9000000000000001 J

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3 years ago

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Answer:

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