Please Help!! This is very tough for me. Try you best to answer these problems!!Pleaseee

A ball falling through the air has what?

AysviL [449]

A ball falling through the air has a mass, a density, a volume...it is facing air resistance and is being acted on by gravity...it is accelerating and gaining velocity...and it is increasing in kinetic energy.
I suppose out of all those the biggest thing the ball has in this case is ENERGY. There are two main types to focus on...

Kinetic Energy - The further the ball fall the more KE it has...until terminal velocity is reach, then KE would become constant.
Potential Energy - Conversely to that of KE, the further the ball falls the less PE it will have.

Heat/Thermal Energy is technically also present due to the friction from the air resistance, but the transfer of energy between the air and ball is quite complex and not necessary important for basic physics.

The question itself seem kind of vague and open ended, but I could just be viewing it the wrong way.
Comment if you need more help!

8 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

For the course of upward acceleration:

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

Now the velocity of the rocket just after the fuel is finished:

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago

A stone is launched from the ground, at a 70° angle, with an initial velocity of 120 m/s.

zavuch27 [327]

A) x = 41t
The classic equation for distance is velocity multiplied by time. And unfortunately, all of your available options have the form of that equation. In fact, the only difference between any of the equations is what looks to be velocity. And in order to solve the problem initially, you need to divide the velocity vector into a vertical velocity vector and a horizontal velocity vector. And the horizontal velocity vector is simply the cosine of the angle multiplied by the total velocity. So H = 120*cos(70) = 120*0.34202 = 41.04242 So the horizontal velocity is about 41 m/s. Looking at the available options, only "A" even comes close.

3 0

3 years ago

Read 2 more answers

This is a bond in which a single pair of electrons is shared between a pair of atoms.

Murrr4er [49]

Answer:single bond

Explanation:

8 0

3 years ago

During a race, four competitors of the same weight rode identical bicycles for 10 minutes. At 8 minutes, which bicycle was movin

Sophie [7]

Answer:

All the competitors will move with the same velocity.

Explanation:

Here, the situations for each competitor are identical. Thus, they will exert the same force and hence, their velocities at each instants will be identical.

6 0

3 years ago