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3 years ago

A 25 kg rock falls for 2.7 seconds. How many meters tall was the

Physics

1 answer:

jek_recluse [69]3 years ago

3 0

Answer:

35.72 m

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 2.7 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (2.7 s) + ½ (9.8 m/s²) (2.7 s)²

Δx = 35.72 m

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A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4. 5 m/s. What is the car’s change in momentum d

kodGreya [7K]

Answer:

Explanation:

Change in car's momentum = 700 * [4.5 - {-29)] = 23,450 kgm/s

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3 years ago

why do signs at petrol stations warn people not to use cell phones while filling the petrol in their cars?

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Cell phones may be radioactive.Maybe this is the reason???.

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3 years ago

For a wave, the _____ the amplitude, the _____ energy the wave carries.

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Answer:

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Explanation:

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3 years ago

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Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of

sp2606 [1]

Answer:

Explanation:

Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg

Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²

For change in period of rotation we shall apply conservation of angular momentum law

I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .

I₁ / I₂ = ω₂ / ω₁

I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .

T₂ / T₁ = I₂ / I₁

(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²

1 + 5 / 3 x 2.3 x 10¹⁹ / M

= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴

= 1 + .0000064

T₂ = 24 (1 + .0000064)

= 24 hours + .55 s

change in length of the day = .55 s .

3 0

3 years ago

A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al

anzhelika [568]

Answer:

aₓ = 0 , ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

vₓ = v₀ₓ = 1.10 m / s

aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

$v_{y}$ = v_{oy} -ay t

ay = (v_{oy} -v_{y}) / t

ay = (0 -10.9) / 1.6

ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

8 0

3 years ago