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Akimi4 [234]
3 years ago
14

how could you measure the amount of elastic potential energy in a streched rubber band? think about the definition of energy whe

n planning your investigation
Physics
1 answer:
irina1246 [14]3 years ago
6 0
<h2>The elastic potential energy gained by the rubber band is the work done by you in elongating it. Positive work is done on the rubber band, so it gains energy. While work is done by you (i.e. negative work is done on you), so you lose some energy. The total energy is conserved. </h2><h2> </h2><h2>In simple words, the energy gained by the rubber band is the energy lost by you while elongating it.</h2>
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Which of the following is evidence that supports the idea of uniformitarianism?
Inga [223]
D. rates of soil erosion are much lower during droughts that last several years
5 0
3 years ago
you ride your bike for a distance of 30km. you travel ata a speed of 0.75km/minute.how many minutes does it take
mihalych1998 [28]
Distance for which the bike is ridden = 30 km
Speed at which the bike is driven = 0.75 km/minute
Let us assume the number of minutes taken to travel the distance of 30 km = x
Now we already know the formula of speed can be written as
Speed = Distance traveled/ Time taken
0.75 = 30/x
0.75x = 30
x = 30/0.75
  = 40 minutes
So the time taken for riding a distance of 30 km will be 40 minutes. I hope this procedure is simple enough for you to understand.
7 0
3 years ago
Read 2 more answers
Three capacitors having capacitances of 8.40, 8.40, and 4.20 μFμF, respectively, are connected in series across a 36.0-V potenti
son4ous [18]

Answer:

a)Q=71.4 μ C

b)ΔV' = 10.2 V

Explanation:

Given that

C ₁= 8.7 μF

C₂ = 8.2 μF

C₃ = 4.1 μF

The potential difference of the battery, ΔV= 34 V

When connected in series

1/C = 1/C ₁ + 1/C₂ + 1/C₃

1/ C= 1/8.4 +1 / 8.4 + 1/4.2

C=2.1 μF

As we know that when capacitor are connected in series then they have same charge,Q

Q= C ΔV

Q= 2.1 x 34 μ C

Q=71.4 μ C

b)

As we know that when capacitor are connected in parallel then they have same voltage difference.

Q'= C' ΔV'

C'= C ₁+C₂+C₃        (For parallel connection)

C'= 8.4 + 8.4 + 4.2 μF

C'=21 μF

Q'= C' ΔV'

Q'=3 Q

3 x 71.4= 21 ΔV'

ΔV' = 10.2 V

3 0
3 years ago
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the s
Nostrana [21]

<u><em>The  question doesn't provide enough data to be solved, but I'm assuming some magnitudes to help you to solve your own problem</em></u>

Answer:

<em>The maximum height is 0.10 meters</em>

Explanation:

<u>Energy Transformation</u>

It's referred to as the change of one energy from one form to another or others. If we compress a spring and then release it with an object being launched on top of it, all the spring (elastic) potential energy is transformed into kinetic and gravitational energies. When the object stops in the air, all the initial energy is now gravitational potential energy.

If a spring of constant K is compressed a distance x, its potential energy is

\displaystyle P_E=\frac{Kx^2}{2}

When the launched object (mass m) reaches its max height h, all that energy is now gravitational, which is computed as

U=mgh

We have then,

U=P_E

\displaystyle mgh=\frac{Kx^2}{2}

Solving for h

\displaystyle h=\frac{Kx^2}{2mg}

We have little data to work on the problem, so we'll assume some values to answer the question and help to solve the problem at hand

Let's say: x=0.2 m (given), K=100 N/m, m=2 kg

Computing the maximum height

\displaystyle h=\frac{(100)0.2^2}{2(2)(9.8)}

\displaystyle h=\frac{4}{39.2}=0,10\ m

The maximum height is 0.10 meters

8 0
2 years ago
Two cars, one in front of the other, are traveling down the highway at 25 m/s. the car behind sounds its horn, which has a frequ
netineya [11]
<span>You are given two cars, one in front of the other, that are traveling down the highway at 25 m/s. You are also given a frequency of 500 Hz of the car travelling behind it. You are asked what is the frequency heard by the driver of the lead car. This problem can be solved using the Doppler effect

sound frequency heard by the lead car = [(speed of sound + lead car velocity)/( speed of sound + behind car velocity)] * (sound of frequency of the behind car)
</span>sound frequency heard by the lead car = [(340 m/s + 25 m/s)/(340 m/s - 25 m/s)] * (500 Hz)
sound frequency heard by the lead car = 579 Hz
7 0
3 years ago
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