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Elis [28]
2 years ago
8

A student is observing a pendulum swinging back and forth. Each time it swings, the pendulum bob reaches a lower maximum, and ev

entually it comes to rest and hangs straight down vertically. What happens to the total energy of the pendulum? Does this violate the law of conservation of energy? Explain your reasoning.
Physics
2 answers:
vodomira [7]2 years ago
7 0

Answer:

the pendulum loses momentum and stops because of gravity and wind resistance. it does not violate the law of conservation of energy because it is not gaining any more momentum than what it had started with

Explanation:

RSB [31]2 years ago
6 0
The total energy of the pendulum stays the same. This does not violate the law of conversation of energy because that law states energy cannot be created/destroyed and the total amount of energy has stayed the same.
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When sugar or another substance is dissolved in water, it disappears from view and forms a homogeneous mixture with the water, a
adoni [48]
You can tell if the sugar is still there by boiling off the water and leaving the sugar behind in the container. Sugar is a solid, and therefore cannot evaporate, so when the water reaches boiling point, it will evaporate at a quicker rate than before (water evaporates at any temperature in liquid form; just not enough to be noticeable) and leave the container to become water vapour
6 0
3 years ago
A car accelerates uniformly from rest and reaches a speed of 22.5 m/s in 8.96 s. Assume the diameter of a tire is 58.9 cm. (a) F
Murljashka [212]

Answer:

54.4747649021

38.2003395586 rad/s

Explanation:

Acceleration

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{22.5-0}{8.96}\\\Rightarrow a=2.51116071429\ m/s^2

Distance covered

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 2.51116071429\times 8.96^2\\\Rightarrow s=100.8\ m

Number of revolutions

n=\dfrac{s}{\pi d}\\\Rightarrow n=\dfrac{100.8}{\pi 0.589}\\\Rightarrow n=54.4747649021

Number of revolutions is 54.4747649021

Angular speed is given by

\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{22.5}{0.589}\\\Rightarrow \omega=38.2003395586\ rad/s

The final angular speed is 38.2003395586 rad/s

8 0
4 years ago
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim
Schach [20]

Complete Question

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn.

Required:

a. What is the race car's centripetal (radial) acceleration?

b. What is the force responsible for the centripetal acceleration in this case?

O normal

O gravity

O friction

O weight

Answer:

question a

       a = 16.32 \  m/s^2

question b

        correct option is option 3

Explanation:

From the question we are told that

   The radius is  r = 57.0 \ m \

    The constant speed at which the race car is travelling is v  = 30 .5 \ m/s

Generally from the question we are told that the track is completely flat so the only force pulling the car to the middle is the frictional force hence the centripetal force is due to the frictional force

    Generally the centripetal acceleration is mathematically represented as

      a = \frac{v^2}{r}

=>    a = \frac{30.5^2}{ 57}

=>    a = 16.32 \  m/s^2

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I believe the answer is free electrons
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What would an automobile engineer MOST LIKELY suggest to improve the efficiency of a car’s engine?
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The answer is only yes cause they try to make it run more efficient

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