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Pepsi [2]
3 years ago
9

A 7.40-kg object initially has 347 J of gravitational potential energy. Then an elevator lifts the object a distance of 20.6 m a

bove its previous position. How much work did the elevator perform on the object?
Physics
1 answer:
storchak [24]3 years ago
6 0

Answer:

W = 1493.9 J = 1.49 KJ

Explanation:

The work done by the elevator on the object will be equal to the gain in is potential energy:

W = ΔP.E

W = mgΔh

where,

W = Work = ?

m = mass of object = 7.4 kg

g = 9.8 m/s²

Δh = gain in height = 20.6 m

Therefore,

W = (7.4 kg)(9.8 m/s²)(20.6 m)

<u>W = 1493.9 J = 1.49 KJ</u>

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You are driving at 40 miles per hour and you speed up to 60 miles per hour over the course of 20 miles. How long were
notka56 [123]

Answer:

120 hours

Explanation:

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3 years ago
Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.
rjkz [21]

Answer:

Momentum of block B after collision =-50\ kg\ ms^{-1}

Explanation:

Given

Before collision:

Momentum of block A = p_{A1}= -100\ kg\ ms^{-1}

Momentum of block B = p_{B1}= -150\ kg\ ms^{-1}

After collision:

Momentum of block A = p_{A2}= -200\ kg\ ms^{-1}

Applying law of conservation of momentum to find momentum of block B after collision p_{B2}.

p_{A1}+p_{B1}=p_{A2}+p_{B2}

Plugging in the given values and simplifying.

-100-150=-200+p_{B2}

-250=-200+p_{B2}

Adding 200 to both sides.

200-250=-200+p_{B2}+200

-50=p_{B2}

∴ p_{B2}=-50\ kg\ ms^{-1}

Momentum of block B after collision =-50\ kg\ ms^{-1}

6 0
2 years ago
PLSSS HELP I WILL GIVE U BRAINLIEST annabelle and jose are building a complete circuit for their science inquiry in class they w
Agata [3.3K]

Answer:

i think its C hope this is rhight have a good day

Explanation:

6 0
3 years ago
Read 2 more answers
A 2kg blob of putty moving at 4 m/s slams into a 6kg blob of putty at rest what is the speed of the to stick together blobs imme
andrew11 [14]
<h2>Answer</h2>

1m/s

<h2>Explanation</h2>

Given that:

<em>Mass of first blob = 2kg = m1</em>

<em>Velocity of blob = 4m/s = v1</em>

<em>Mass of second blob = 6kg = m2</em>

<em>Velocity of blob = 0m/s = v2</em>

<em />

To find:

<em>Final velocity = Vf</em>

<em />

<em>This question is of inelastic collision which is any collision between objects in which some energy is lost.</em>

<em />

<h3>Formula to be use:</h3><h2>(m1*v1) + (m2*V2) = Vf(m1 + m2)</h2>

(2*4) + (6*0) = Vf(2+6)

8 + 0 = Vf(8)

8 = Vf(8)

Vf = 1 m/s

So the speed of two blobs immediately after colliding = 1 m/s

3 0
3 years ago
Light in a vacuum travels at a constant speed of 3x10^8 m/s. If the moons average distance from the earth is 38776106 km how lon
Karo-lina-s [1.5K]

Answer: 258.3 s

Explanation:

The speed s is given by the following equation:

s=\frac{D}{t}

Where:

s=3(10)^{8} m/s is the speed of light in vacuum

D=2(38776106 km \frac{1000 m}{1 km})=7.75(10)^{10} m is the double of the distance between Earth and Moon, since the beam of light travels from Earth to the Moon and back to Earth again.

t is the time it takes to the beam of light to travel the mentioned distance

Isolating t and solving with the given information:

t=\frac{D}{s}

t=\frac{7.75(10)^{10} m}{3(10)^{8} m/s}

Finally:

t=258.3 s \approx 258 s

5 0
3 years ago
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