Ask question

Ask question

All categories

3 years ago

8

A student claims "everything falls at the same acceleration rate on the Moon, where there is no air or friction," how would you

respond to that student?

1 answer:

Alexeev081 [22]3 years ago

5 0

Now let’s say you’re on the Moon. If you were to drop a hammer and a feather from the same height, which would hit the ground first?

Trick Question! On the moon both objects would hit the ground at the same time. On Earth, the hammer lands first.

So yeah, the student is right. Galileo gave us this theory long ago.

You might be interested in

Consider this situation: A rope is used to lift an actor upward off the stage. Of the forces listed, identify which act upon the

Katarina [22]

ytyjjryjtjtyjtyjtyyjetyeyjetyyjeytjtyjetyjtjetyjetyeyjetyj

6 0

4 years ago

Read 2 more answers

A student claims an object in motion must experience a force to stay in motion. Do you agree or disagree?

Vikki [24]

Answer:

agree because there is always a force that causes motion..

4 0

3 years ago

Use this technique to find a formula for the intensity I of a sound, in terms of the sound level β and the reference intensity I

mezya [45]

The problem is basically asking us to find a way to find the sound intensity I, in terms dependent on the sound level and the reference intensity $I_0$ .For this purpose we can start from the unit used in the scale logarithmic decibel, that is

$\beta = 10log_{10}\frac{I}{I_0}$

Where

I = Acoustic intensity on the linear scale

$I_0 =$ Hearing threshold

Using the logarithmic properties of the exponents the above expression can be described as:

$(\frac{I}{I_0})^{10} = 10^{\beta}$

$I = I_0 10^{\frac{\beta}{10}} \righarrow$ that is the expression or technique to find the intensity of sound.

8 0

3 years ago

Due Ma<br> duart<br> ded<br> out<br> 25 N<br> 35 N<br> 1-03

Anuta_ua [19.1K]

Answer:

what's that all about

hehehwhe

Explanation:

dgbjjjedgkigdssfhkkoyddwrhkoyeqaxghjjhasghffhjiopjtewqetujjgda

8 0

3 years ago

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

<u>Time taken to descent the total height:</u>

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

<u>Now the total time taken by the ball to hit the ground:</u>

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago