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Vlad [161]
3 years ago
6

A scientist is studying a shock wave from an earthquake. What kind of wave is being studying?

Physics
1 answer:
Pavel [41]3 years ago
3 0

Answer:

Longitudinal Mechanical Wave

Explanation:

Mechanical waves are the waves that require medium to propagate. And a longitudinal wave is a wave in which the vibration of the energy(here: mass specifically) is in the direction of propagation of wave.

Shock wave, strong pressure wave in any elastic medium such as air, water, or a solid substance, produced by supersonic aircraft, explosions, lightning, or other phenomena that create violent changes in pressure.

Shock waves travel faster than sound and their speed increases as the amplitude of the wave is increased but their intensity fades faster due to the fact that some of its energy gets expended in the form of heat due to the resistance of the medium.

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A pulley is most closely related to which simple machine?
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Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object
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Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem.  Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them.  So let's call them F1 and F2, with F1 arger than F2.  Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.  

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A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the
Mademuasel [1]

Answer:

(a) a=2m/sec^2

(b) 5220 j

(c) 1740 watt

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Explanation:

We have given mass m = 290 kg

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Time t = 3 sec

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v = u+at

So a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2

(a) We know that force is given by

F = ma

So force will be F=290\times 2=580N

(b) From second equation of motion we know that

s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m

We know that work done is given by

W = F s = 580×9 =5220 j

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P=\frac{W}{t}=\frac{5220}{3}=1740Watt

(d) Time t = 1.5 sec

So P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt

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