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3 years ago

A tennis ball is dropped from 1.65 m above

Physics

1 answer:

dexar [7]3 years ago

7 0

Answer:

The velocity with which the ball strikes the ground = -5.7 m/s

Explanation:

To find the velocity with which the tennis ball hits the ground, we only need to worry about what happens up to that point. We can ignore the rebound for this part. Given:

d = -1.65

a = -9.8

vi = 0

vf = ?

$v_f^2 = v_i^2 + 2ad\\v_f^2=0+2(-9.8)(-1.65)\\v_f^2=32.34\\v_f=-5.7$

*Keep in mind that the square root gives us two answers, a positve and a negative one. We use the negative one here because the final speed is downwards and the question says down is negative.

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So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

$cos \theta = \frac{AS}{h}$

In this case:

$cos \theta = \frac{r}{R_{E}}$

so we can solve for r, so we get:

$r= R_{E}cos \theta$

in this case we'll use the average radius of earch which is 6,371 km, so we get:

$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

$a=24.07 mm/s^{2}$

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2 years ago

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 108 m. Find

Goshia [24]

Answer:

It took 1.28 seconds to his voice to reach the Earth via radio waves.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities, that radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies, the highest being that of gamma rays, followed by X-rays, ultraviolet rays and the visible region , and those of lower frequencies, which correspond to infrared, microwave and radio waves.

Light propagates as electromagnetic wave in vacuum with a speed of $3x10^{8}m/s$ . Therefore, radio waves will have in vacuum the same speed.

Then, to know the time that it took for its voice, the next equation can be used:

$c = \frac{d}{t}$ (1)

Where c is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

$t = \frac{d}{c}$ (2)

$t = \frac{3.85x10^{8} m}{3x10^{8}m/s}$

$t = 1.28s$

Hence, it took 1.28 seconds to his voice to reach the Earth via radio waves.

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