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Alja [10]
3 years ago
15

A compression wave of gas, liquid, or solid, that is detected by your ear.

Physics
2 answers:
Serggg [28]3 years ago
3 0
Sound 

is a compression wave of gas, liquid, or solid, that is detected by your ear.
creativ13 [48]3 years ago
3 0

Answer:

sound

Explanation:

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What are Newton's 1st 2nd and 3rd laws of motion?

They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. More precisely, the first law defines the force qualitatively, the second law offers a quantitative measure of the force, and the third asserts that a single isolated force doesn't exist.

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They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. More precisely, the first law defines the force qualitatively, the second law offers a quantitative measure of the force, and the third asserts that a single isolated force doesn't exist.

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Read 2 more answers
Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/
IrinaK [193]

Answer: (a) 1.065 N  (b) 2.13 N

Explanation:

<h2>(a) average force exerted by the rain on the roof</h2><h2 />

According Newton's 2nd Law of Motion the force F is defined as <u>the variation of linear momentum</u> p <u>in time:</u>

F=\frac{dp}{dt}  (1)

Where the linear momentum is:

p=mV  (2) Being m the mass and V the velocity.

In the case of the rain drops, which initial velocity is V_{i}=15m/s and final velocity is  V_{f}=0 (we are told the drops come to rest after striking the roof). The momentum of the drops p_{drops} is:

p_{drops}=mV_{i}+mV_{f}  (3)

If V_{f}=0, then:

p_{drops}=mV_{i}  (4)

Now the force F_{drops} exerted by the drops is:

F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}  (5)

F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}  (6)

At this point we know the mass of rain per second (mass rate) \frac{dm}{dt}=0.071 kg/s and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

F_{drops}=\frac{dm}{dt}V_{i}  (7)

F_{drops}=(0.071 kg/s)(15m/s)  (8)

F_{drops}=1.065kg.m/s^{2}=1.065N  (9) This is the force exerted by the rain drops on the roof of the car.

<h2>(b) average force exerted by hailstones on the roof </h2><h2 />

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum p_{hailstones} is:

p_{hailstones}=mV_{i}+mV_{f}   (9)  Being V_{i}=V_{f}

p_{hailstones}=mV+mV=2mV   (10)  

Deriving with respect to time to find the force F_{hailstones} exerted by the hailstones:

F_{hailstones}=\frac{d}{dt}p_{hailstones}=\frac{d}{dt}(2mV)   (10)  

F_{hailstones}=2\frac{d}{dt}(mV)=2(\frac{dm}{dt}V+m\frac{dV}{dt})   (11)  

Assuming \frac{dV}{dt}=0:

F_{hailstones}=2(\frac{dm}{dt}V)   (12)  

F_{hailstones}=2(0.071 kg/s)(15m/s)   (13)  

Finally:

F_{hailstones}=2.13kg.m/s^{2}=2.13N (14)   This is the force exerted by the hailstones  

Comparing (9) and (14) we can conclude the force exerted by the hailstones is two times greater than the force exerted by the raindrops.

5 0
3 years ago
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