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nasty-shy [4]
3 years ago
14

What is the potential energy of a 2 kg ball 15 m in the air?

Physics
1 answer:
oee [108]3 years ago
7 0

Answer:

<h2>300 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 2 × 10 × 15

We have the final answer as

<h3>300 J</h3>

Hope this helps you

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A bungee jumper attains a speed of 30 m/s just as the bungee cord begins to stretch. If the period of stretch is 2 s while comin
Daniel [21]

Answer:

1.53 g

Explanation:

Given,

Initial speed = 30 m/s

Final speed = 0 m/s

Period of stretch, t = 2 s

average deceleration = ?

we know

a = \dfrac{v-u}{t}

a = \dfrac{0-30}{2}

a = -15 m/s²

Deceleration of the jumper = 15 m/s²

Deceleration in terms of g

= \dfrac{15}{9.81}

= 1.53\ g

Hence, the deceleration of the jumper is equal to 1.53 g

7 0
3 years ago
The uncertainty Δp sets a lower bound on the average momentum of a particle in the nucleus. If a particle's average momentum wer
kirill [66]

Answer: The minimum kinetic energy Kmin is 1.3 × 10^-13 J

Explanation: Please see the attachments below

8 0
3 years ago
M
Zigmanuir [339]

Answer:

it’s 2.5 m/s

Explanation:

i’m too lazy but trust

6 0
2 years ago
Consider two spaceships, each traveling at 0.50c in a straight line. Ship A is moving directly away from the Sun and ship B is a
attashe74 [19]

Answer:

The velocity of the light will be 1.0c only

Explanation:

The velocity of the light measured in the case given in question will be 1.0c only.

This is due to the fact that the velocity of light is never relative. The velocity of the light is maximum

The velocity of the light cannot be scaled down in no case

Thus, the velocity of the light remains as constant.

Hence, the velocity of the light measured will be 1.0c although the ships have relative velocity.

3 0
2 years ago
Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s
ozzi

Answer:

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe       ..... (1)

T Cosθ = mg     .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}

tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}

tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}

tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}

As θ is very small, so tanθ and Sinθ is equal to θ.

\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

7 0
3 years ago
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