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katrin2010 [14]
3 years ago
14

If it requires 5.0 j of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be re

quired to stretch it an additional 3.7 cm ?
Physics
1 answer:
ANEK [815]3 years ago
5 0
15.277.. j. I did the problem using a proportion. an additional 3.7m to the current 1.8 cm=5.5cm.

Therefore, 5.0 j/1.8cm=x/5.5cm
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An object accelerates 3 m/s2 , when a force of 6 N acts on it. What is the object’s mass
telo118 [61]

Answer:

<h2>2 kg</h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{6}{3}  \\  = 2

We have the final answer as

<h3>2 kg</h3>

Hope this helps you

3 0
3 years ago
A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo
cestrela7 [59]

Answer:

Volume of balloon =  1000 cm^3

Explanation:

 The head of a normal person can be assumed as a sphere with radius 10 cm.

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 We have approximate radius = 10 cm.

  Approximate volume of head =\frac{4}{3} \pi r^3=\frac{4}{3} *\pi* 10^3=4188cm^3

 In the given options the closest value to the approximate volume is 1000 cm^3.

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4 0
3 years ago
A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.
brilliants [131]
40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters
7 0
4 years ago
A bowling ball rolled with a force of 20 N accelerates at a rate of 5 m/sec2, a second ball rolled with the same force accelerat
emmasim [6.3K]

Answer:

sorry I need points

Explanation:

IM SO SORRY

5 0
3 years ago
The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this
antiseptic1488 [7]

Answer:

<h3>14.97m/s</h3>

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

v² = 64+160

v² = 224

v = √224

v = 14.97m/s

Hence the rider's velocity after the acceleration is 14.97m/s

5 0
3 years ago
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