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3 years ago

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A common acoustic phenomenon is thunder, the rapid heating and supersonic expansion of air created by a lightning stroke. Becaus

e the speed of light reaches almost any point on Earth instantly while sound travels more slowly, the time lag between lightning and thunder can be used to estimate distance to the lightning. This delay time is ______ seconds per kilometer (5 sec/mi).

1 answer:

Oksana_A [137]3 years ago

4 0

Answer:

Δt ≈ 2.9137 ≈ 3 seconds per kilometer

Explanation:

The speed of light is approximately 300000 km /s, while the speed fo the sound in the air is 0.3432 km/s.

The light takes therefore this time to travel one kilometer

$x = vt$ $t = \frac{x}{v}$

$t = \frac{1}{300000}$

$t = 3.33 x 10^{-6} s$

On the other hand the sound takes this time to travel one kilometer

$t = \frac{1}{0.3432}$

t = 2.9137 s

Then the delay time is 2.9137 - $3.33 x 10^{-6} s$

Δt ≈ 2.9137 ≈ 3 s

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. A 2.0-kg block is on a perfectly smooth ramp that makes an angle of 30° with the horizontal. (a) What is the block’s accelerat

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Answer:

a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N

Explanation:

a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry

sin 30 = Wx / W

cos 30 = Wy / W

Wx = W sin30

Wy = W cos 30

Let's write the equations on each axis

X axis

Wx = ma

Y Axis

N- Wy = 0

N = Wy = mg cos 30

N = 2.0 9.8 cos 30

N = 16.97 N

We calculate the acceleration

a = Wx / m

a = mg sin 30 / m

a = g sin 30

a =9.8 sin 30

a = 4.9 m / s²

b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component

F -Wx = 0

F = Wx

F = m g sin 30

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In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

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