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Nadya [2.5K]
3 years ago
7

At what height above the ground must a mass of 10 kg be to have a potential energy equal in value to the kinetic energy possesse

d by a mass of 10 kg moving with a velocity of 20 m/s? 100 m 10 m 50 m 20 m
Physics
1 answer:
Paladinen [302]3 years ago
5 0

Answer:

20 m

Explanation:

We'll begin by calculating the kinetic energy of the mass. This can be obtained as follow:

Mass (m) = 10 kg

Velocity (v) = 20 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 10 × 20²

KE = 5 × 400

KE = 2000 J

Finally, we shall the height to which the mass must be located in order to have potential energy that is the same as the kinetic energy. This can be obtained as follow:

Mass (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s²

Potential energy (PE) = Kinetic energy (KE) = 2000 J

Height (h) =..?

PE = mgh

2000 = 10 × 10 × h

2000 = 100 × h

Divide both side by 100

h = 2000 / 100

h = 20 m

Thus, the object must be located at a height of 20 m in order to have potential energy that is the same as the kinetic energy.

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Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb
Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

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Before the leakage, the tire walls are kept tensed due to the internal pressure exerted by the air molecules. When the leakage occurs the air molecules escape and this makes the tire loose its firm upright shape thereby falling to the floor and creating an increase in the area of contact between the tire and the floor surface.

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