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3 years ago

The law of conservation of energy stated that ?

Physics

1 answer:

givi [52]3 years ago

5 0

Answer:

The total energy of an isolated remains constant. It is said to be conserved over time.

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6/10

g100num [7]

Answer:pounds

Explanation:

7 0

3 years ago

The distance between a speaker and a listener is 8.0 m. Determine the frequency of the sound wave if exactly 20 waves are formed

dmitriy555 [2]

The frequency of the wave is 6800 Hz

<u>Explanation:</u>

Given:

Wave number, n = 20

Speed of light, v = 340 m/s

Frequency, f = ?

we know:

wave number = $\frac{frequency}{speed of light}$

$20 = \frac{f}{340 m/s} \\\\f = 20 X 340 s^-^1\\\\f = 6800 Hz$

Therefore, the frequency of the wave is 6800 Hz

4 0

3 years ago

18. If the distance from a converging lens to the object is less than the focal length of the lens, the image will be _______ th

skad [1K]

Here's a quick way to find out. Pick up your glasses, bifocals work best, and find the focal length with a flashlight against a book. If I remember right, the object should be magnified and upside down. So, A.

5 0

3 years ago

-What do you think happened to make the Moon look the way it does?

goldenfox [79]

Answer: The physics of evolution had made the moon like it is today....Please watch this video from you tube about the evolution of the moon.

Explanation:

https://youtu.be/UIKmSQqp8wY

5 0

3 years ago

A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th

Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

$Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

now we have

$W_{gravity} = -mg(Lsin\theta)$

$W_{friction} = -\mu mgcos(\theta) L$

so from above equation

$Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

so from above equation work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

8 0

3 years ago