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icang [17]
2 years ago
6

A 1000 N crate is lifted to a height of 3.0 m. How much work is done to lift the crate?

Physics
1 answer:
Bogdan [553]2 years ago
5 0

Answer:

0

Explanation:

o

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The answer is moderate intensity
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Can i get help for the parts for the two questions? MATHPHYSSSS
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Explanation:

003 (part 1 of 2)

Pressure is force divided by area.

P = F / A

P = (117 kg × 9.8 m/s²) / (2 × (0.05 m)²)

P = 229,320 Pa

003 (part 2 of 2)

There are approximately 6895 Pa in 1 psi.

P = 229,320 Pa × (1 psi / 6895 Pa)

P = 33.3 psi

004 (part 1 of 2)

Since the collisions are elastic, the angle of reflection is the same as the angle of incidence (it bounces off at the same angle).

Impulse = change in momentum

F Δt = m Δv

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F = 0.218 N

004 (part 2 of 2)

Pressure is force over area.

P = F / A

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7 0
3 years ago
Nitrogen is used to proctect steel from water because nitrigen has?
blondinia [14]
<span>Nitrogen is used to protect steel from water because nitrogen has stable physical properties that enables the nitrogen to be resistant to water molecules. Nitrogen also serves as a coat for the steel to not be corroded easily</span>
8 0
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A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate
natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

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2 years ago
A rock is dropped from the top of a 500m cliff. How long does it take the rock to reach the bottom
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5 seconds.............
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