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3 years ago

A ball is thrown vertically upward with a speed of 12.0 m/s. ( a. How high does it rise?

1 answer:

5 0

It rises till all of its Kinetic energy is converted into potential energy.

so, mgh=(1/2)m(v^2)

so, h=(v^2)/2g = 12*12/(2*9.81)=7.34 m

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andriy [413]

Answer:

Explanation:

It will stay in motion unless for example, someone stops it from going.

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3 years ago

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A ball is dropped from somewhere above a window that is 2.00 m in height. As it falls, it is visible to a person boxing through

Irina18 [472]

Answer:

4.14 m

Explanation:

In the last leg of the journey the ball covers 2 m in 2ms or 0.2 s .

Let in this last leg , u be the initial velocity.

s = ut + 1/2 g t²

2 = .2 u + .5 x 9.8 x .04

u = 9.02 m /s .

Let v be the final velocity in this leg

v² = u² + 2 g s

v² = (9.02)² + 2 x 9.8 x 2

= 81.36 +39.2

v = 10.97 m / s

Now consider the whole height from where the ball dropped . Let it be h.

Initial velocity u = 0

v² = u² +2gh

(10.97 )² = 2 x 9.8 h

h = 6.14 m

Height from window

= 6.14 - 2m

= 4.14 m

5 0

4 years ago

A 0.25 kg book falls off a 2 m shelf on to a 0.5 m chair. What was the change in GPE?

Snowcat [4.5K]

The equation of GPE is mgH, where m is mass, g is gravitational acceleration, and H is the height.

If we're solving for the change in GPE, then:

∆ $U_{g}$ = mg∆H

Input our given values for m and g:

∆ $U_{g}$ = 0.25 * 9.80 * ∆H

The book falls from 2 meters high to 0.5 meters high, so:

∆ $U_{g}$ = 0.25 * 9.80 * (2.0 - 0.5)

∆ $U_{g}$ = 0.25 * 9.80 * 1.5

∆ $U_{g}$ = 3.675 (J)

Adjust for significant figures:

∆ $U_{g}$ = 3.7 (J)

The change in gravitational potential energy was 3.7 (J)

If you have any questions on anything I did to get to the answer, just ask!

- breezyツ

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3 years ago

Calculate the electric field at the center of a square 46.4 cm on a side, if one corner is occupied by a +42.0 µc charge and the

liraira [26]

centre of square disrance to each corner found by Pythagoras' theorem.

coulombs law used to clculate field of each charge at centre

fields added vectorially for res

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3 years ago

If the mass of a planet is 0.231 mE and its radius is 0.528 rE, estimate the gravitational field g at the surface of the planet.

crimeas [40]

Answer:

$8.1 m/s^2$

Explanation:

The strength of the gravitational field at the surface of a planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For the Earth:

$g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2$

For the unknown planet,

$M_X = 0.231 M_E\\R_X = 0.528 R_E$

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

$g_X = \frac{GM_X}{R_X^2}=\frac{G(0.231M_E)}{(0.528R_E)^2}=\frac{0.231}{0.528^2}(\frac{GM_E}{R_E^2})=0.829 g_E$

And substituting g = 9.8 m/s^2,

$g_X = 0.829(9.8)=8.1 m/s^2$

3 0

3 years ago