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Margaret [11]
3 years ago
10

You raise a bucket of water from the bottom of a well that is 12 m deep. the mass of the bucket and the water is 5.00 kg, and it

takes 15 s to raise the bucket to the top of the well. how much power is required
Physics
1 answer:
MArishka [77]3 years ago
4 0
Given: Mass m = 5.00 Kg;    Height h = 12 m;    Time t = 15 s

Required: Power P = ?

Formula: P = Fd/t  = mgh/t

               P = (5.0 Kg)(9.8 m/s²)(12 m)/15 s

               P = 39.2 Kg.m²/s²  or
 
               P = 39.2 J





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D. Steel is made of different compounds distributed unevenly.

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A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the tables edge in the
GREYUIT [131]

A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the edge of the table in the horizontal direction,

  • t= 0.45 seconds.
  • V=2.22m/s
  • VT=4.95 m/s

This is further explained below.

<h3>What is its speed when it hits the floor...?</h3>

Generally, the equation for motion is mathematically given as

S= ut + 0.5at²

Therefore

y = Voy t + 0.5gt^2

1 = 0.5x 98 x 6²

1=4.9t^2

t=\sqrt{0.2041 }

t= 0.45 seconds.

b) Horizontal motions are uniform.

V=Horizontal displacement/time

V=1/0.45

V=2.22m/s

C)

Vx: 2.22 m/s At bottom,

Vy² = Voy² + 2as

Vy² = 2x95x1

Vy² = 19.6

Total velocity

VT=\sqrt{( 2.22 m/)^2+19.6}

VT=4.95 m/s

Read more about  Arithmetic

brainly.com/question/22568180

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4 0
2 years ago
A motor must lift a 1500-kg elevator cab. The cab's maximum occupant capacity is 400 kg, and its constant "cruising" speed is 1.
natka813 [3]

Answer:

12900 W

24200 W

Explanation:

Given:

v₀ = 0 m/s

v = 1.3 m/s

t = 2.0 s

Find: a and Δx

v = at + v₀

(1.3 m/s) = a (2.0 s) + (0 m/s)

a = 0.65 m/s²

Δx = ½ (v + v₀) t

Δx = ½ (1.3 m/s + 0 m/s) (2.0 s)

Δx = 1.3 m

While accelerating:

Newton's second law:

∑F = ma

F − mg = ma

F = m (g + a)

F = (1500 kg + 400 kg) (9.8 m/s² + 0.65 m/s²)

F = 19855 N

Power = work / time

P = W / t

P = Fd / t

P = (19855 N) (1.3 m) / (2.0 s)

P ≈ 12900 W

At constant speed:

Newton's second law:

∑F = ma

F − mg = 0

F = mg

F = (1500 kg + 400 kg) (9.8 m/s²)

F = 18620 N

Power = work / time

P = W / t

P = Fd / t

P = Fv

P = (18620 N) (1.3 m/s)

P ≈ 24200 W

5 0
4 years ago
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