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Bumek [7]
3 years ago
8

A 2.0kg object is raised by a height of 0.5m.Calculate the energy raised in its gravitational potential energy store? G=9.8N/KG

Physics
1 answer:
notka56 [123]3 years ago
5 0

Answer:

Gravitational potential energy stored is 9.8 J.

Given:

G = 9.8 \frac{N}{kg}

height = 0.5 m

mass = 2 kg

To find:

Gravitation Potential energy stored = ?

Formula used:

Potential energy = mGh

m = mass

G = acceleration due to gravity

h = height

Solution:

Potential energy of the object is given by,

Potential energy = mGh

m = mass

G = acceleration due to gravity

h = height

Potential energy = 2 × 9.8 × 0.5

Potential energy = 9.8 Joule

Gravitational potential energy stored is 9.8 J.

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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
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Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

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Mass of first broken body, m1= 9.4kg

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Mass of 3rd broken body = ?

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Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

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