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2 years ago

Jane has a mass of 40 kg. She pushes on a rock with a force of 100 N. what force does the rock exert on Jane

Physics

1 answer:

Phoenix [80]2 years ago

5 0

Answer:

100N

Explanation:

if Jane pushes on the rock with 100 Newtons of force, then the rock pushes on Jane with 100 Newtons of force.

hope this helps :)

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What is the momentum of a 5 kg object that has a velocity of 1.2 m/s? 3.8 kg • m/s 4.2 kg • m/s 6.0 kg • m/s 6.2 kg • m/s

Gnesinka [82]

Answer:

Your answer will be 6.0kg•m/s

Explanation:

In the given question all the required details d given. Using these information's a person can easily find the momentum of the object. In the question it is already given that the mass of the object is 5 kg and the velocity at which it is traveling is 1.2 m/s.We know the equation of finding momentum asMomentum = mass * velocity = 5 * 1.2 = 6So the momentum of the object is 6 Newton.

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3 years ago

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MakcuM [25]

The proton would be 2 and d a part of 1 then calculate that hope this helped

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2 years ago

When you cook a marshmallow on a metal poker tool over an open flame, energy is transferred. Identify the three different ways t

sergejj [24]

Answer:

The three ways thermal energy is transferred are;

1) Conduction

2) Convection

3) Radiation

Explanation:

1) The conduction of the heat from the open flame to the marshmallow is through the direct contact of the flame with the marshmallow, such that the flame the region of the combustion reaction, that produces light and heat touches the marshmallow

2) The convection process is the transfer of heat from the rising heated combustion products, as well as the heated air that rises from the flame

3) The radiation heat transfer is the transfer of the heat from the fire to the marshmallows directly by the heat the moves in the form of electromagnetic waves at temperatures above 1000 K, without the need for a medium, such that the marshmallow can be heated by the heat coming from side of the flame.

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3 years ago

Suppose you have a Frisbee with a mass of 0.10 kg. You first throw it with a force of 5N and then change the force you throw it

natali 33 [55]

Answer:

The acceleration increases.

Explanation:

From Newton's 2nd Law, we have $\Sigma F=ma$ . We can see that force is directly proportional to mass and acceleration. Therefore, as force increases, either mass or acceleration must increase as well, and vice versa. Since mass is maintained here, if you increase the force applied to the Frisbee, the acceleration will increase as well.

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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, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

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.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

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:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

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:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

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.

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2 years ago