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SIZIF [17.4K]
3 years ago
15

A rock is dropped from a sea cliff, and the sound of it striking the ocean in heard 3.0.s later. If the speed of sound is 340m/s

, how high is the cliff?​
Physics
1 answer:
Ahat [919]3 years ago
3 0

Answer:

The answer is 1020 meters.

Explanation:

The values given by the problem are:

1. T= The falling time of the rock [Seconds]

2. H=The sound velocity constant [meter/second]

The Velocity normal formula is

V=H/T

340=H/3

340*3=H

This solution considers the physic of the traveling wave sound but not really the rock falling problem, because the problem ask for the height of the cliff not even more.

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Elenna [48]
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4 0
3 years ago
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Pepsi [2]
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5 0
2 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N
Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

4 0
2 years ago
A freight train car has a mass of 2,000 kilograms and an acceleration of 1.8 m/s/s. ​What is the average force behind that train
abruzzese [7]

Answer:

F = 3600 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of force must be equal to the product of mass by acceleration.

ΣF = m*a

where:

F = force [N]

m = mass = 2000 [kg]

a = acceleration = 1.8 [m/s^2]

Now replacing:

F = 2000*1.8

F = 3600 [N]

8 0
3 years ago
A physics book is thrown horizontally at a velocity of 5.0 m/s from the top of a cliff 78.4 m high. How long does the book take
xenn [34]

The book's vertical position in the air is

y=78.4\,\mathrm m-\dfrac12gt^2

where g=9.80\,\frac{\mathrm m}{\mathrm s}. It reaches the ground when y=0, at a time t such that

0=78.4\,\mathrm m-\dfrac12gt^2\implies t=4.00\,\mathrm s

So it takes the book 4 seconds to reach the bottom. The given initial velocity is irrelevant since it only has a horizontal component; vertically, the book is starting from rest.

3 0
3 years ago
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