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ryzh [129]
2 years ago
15

Takes 5 seconds and 2,000 J for a crane to lift a very heavy object. Which of these correctly describes power and energy for thi

s action?
2,000 J of energy was converted into some other energy, requiring 10,000 W of power.
2,000 J of energy was converted into some other energy each second, requiring 400 W of power
2,000 J of power was converted into 400 W of energy.
2,000 J of power was converted into some other power every second, requiring 10,000 W of energy.
Physics
1 answer:
strojnjashka [21]2 years ago
8 0

Answer:

D

Explanation

take 5x2000 and J is the unit for power therfore it cant be A and B and C are not 10,000.

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A round object of mass 10 kg and radius 0.5 m rolls without slipping down a hill from a height of 4.5 m. If its speed at the bot
Mice21 [21]

Answer:

moment of inertia is 2.72 kg m²

Explanation:

given data

mass m = 10kg

height h = 4.5 m

radius r  = 0.5 m

speed  v = 6.5 m/s

to find out

moment of inertia

solution

we apply here conservation of energy

that is

mgh = 1/2 ×mv² + 1/2 × Iω²

here I is moment of inertia we find and

we know ω = Velocity / radius = 6.5 / 0.5 = 13

and g = 9.8

so put here all these value

10 (9.8) 4.5 = 1/2 ×(10)(6.5)² + 1/2 × I(13)²

441 = 211.25 + 1/2 × I( 169 )

I = 2.72

so moment of inertia is 2.72 kg m²

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3 years ago
Do acids or bases turn litmus paper red?
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A Christmas sphere will break if dropped on hardwood floor with a velocity of 2.0 m/s. Mar Estefanía is
Margaret [11]

Answer:

0.17 s

Explanation:

Given:

Δy = 0.15 m

v₀ = 0 m/s

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8 0
3 years ago
A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .
Usimov [2.4K]

The body moves at a velocity of 1.62m/s after the bullet emerges.

<h3>Given:</h3>

Mass of bullet, m_1 = 22g

                               = 0.022 kg

Mass of the block, m_2 = 1.9 kg

Velocity of bullet , v_1 = 265 m/s

v_2 = 0

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

After penetration;

v^{'}_1 =125 m/s

v^{'}_2=?

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

∴ Momentum before = Momentum after

Substitute the given parameters into the formula as shown:

   m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s

Therefore, It moves with a velocity of 1.62 m/s.

Learn more about momentum here:

brainly.com/question/25121535

#SPJ1

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