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expeople1 [14]
4 years ago
15

A 5kg cat is lifted 2m into the air. How much GPE does it gain

Physics
1 answer:
luda_lava [24]4 years ago
5 0

m = mass of the cat being lifted = 5 kg

h = height to which the cat is raised into air = 2 m

g = acceleration due to gravity of earth = 9.8 m/s²

U = gravitational potential energy gained by the cat

Gravitational potential energy gained by the cat is given as

U = m g h

inserting the values in the above equation

U = (5 kg) (9.8 m/s²) (2 m)

U = (5 x 9.8 x 2) (kg m²/s²)

U = 98 J

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Answer:

The displacement can be calculated by measuring the area of the curve when we move along the t curve. So, that would give the displacement.

Explanation:

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What do you need to verify in a patient's profile? A. Name, history of illness, medication, dose, strength, frequency B. Name, a
Lubov Fominskaja [6]
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4 years ago
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7. Two workers must pick up bricks that lie on the ground and place them on a worktable. They each
sdas [7]

Answer:

Same work is done by the two workers

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7 0
4 years ago
Consider two planets of mass m and 2m,
Rzqust [24]

Answer:

Part a)

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Part b)

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Part c)

\frac{T_1}{T_2} = 9.54

Explanation:

Part a)

As we know that the gravitational force is given as

F = \frac{GMm}{r^2}

so we will have to find the ratio of force on two planets due to star

so here we have

\frac{F_1}{F_2} = \frac{m_1r_2^2}{m_2r_1^2}

\frac{F_1}{F_2} = \frac{m (4.5r)^2}{(2m) r}

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Part b)

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v = \sqrt{\frac{GM}{r}}

so the ratio of two orbital speed is given as

\frac{v_1}{v_2} = \frac{r_2}{r_1}

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Part c)

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T = 2\pi\sqrt{\frac{r^3}{GM}}

so the ratio of two time period is given as

\frac{T_1}{T_2} = \sqrt{\frac{r_1^3}{r_2^3}}

\frac{T_1}{T_2} = \sqrt{\frac{4.5r^3}{r^3}}

\frac{T_1}{T_2} = 9.54

8 0
3 years ago
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GaryK [48]

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7 0
4 years ago
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