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kykrilka [37]
3 years ago
7

Which would be the best example to demonstrate the principle of conservation of energy? A) a stone is crushed B) light bends aro

und a sharp edge C) a pendulum swings back and forth D) a balloon inflates when air is blown into it
Physics
2 answers:
Gnom [1K]3 years ago
7 0

The correct answer to the question is : C) A pendulum swings back and forth.

EXPLANATION:

Before answering this question, first we have to understand the law of conservation of energy.

As per law of conservation of energy, energy can neither be created not be destroyed. It can only change from one form to another form, and the total energy of the universe is always constant.

As per the question, we have a pendulum which is moving back and forth.

Let us consider a pendulum which is taken to some height from its equilibrium or mean point. The energy possessed by the pendulum at this height is gravitational potential energy. When the pendulum is released, the potential energy is converted into kinetic energy. At mean point, whole of its potential energy is converted into kinetic energy.

Due to inertia, the pendulum reaches at the other extreme point. During its movement from mean point to extreme point, the kinetic energy is converted into potential energy. At extreme point, whole of its kinetic energy must have converted into potential energy. The same process will be repeated.

Hence, it obeys law of conservation of energy.

Hence, swinging of pendulum back and forth is the best example to demonstrate the law of conservation of energy.

Daniel [21]3 years ago
6 0
The answer is a pendulum swings back an forth.
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A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f
hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2    .... (2)

By equation the equation (1) and (2), we get

41.7=1.12 u +4.9 \times 1.12^{2}

u = 31.75 m/s

5 0
3 years ago
10) If the mass 2m, the left mass
Romashka-Z-Leto [24]

Answer:

F = \frac{-Gm_{1}m_{2} }{r^{2} }.

Explanation:

Gravitational force between two objects of masses m_{1},  m_{2} kept at a distance r is given by the formula

F = \frac{-Gm_{1}m_{2} }{r^{2} }

Here ,m_{1} = 2m

         m_{2} = \frac{m}{2}

         

Thus , F = \frac{-G.2m.\frac{m}{2} }{r^{2} }

          F = \frac{-Gm_{1}m_{2} }{r^{2} }.

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4 0
2 years ago
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PolarNik [594]
Evaporation (or another word to use is water vapor.)
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3 years ago
Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

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e = natural logarithm = 2.718

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V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0
3 years ago
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