F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
F = 56.67N
Answer:
m = 0.111 kg
Explanation:
Heat required to release from the body of the person when his temperature cool down by 1 degree C is given as

here we know that
m = 70 kg
s = 3840 J/kg K

now we know that


now the same heat is used to vaporize water of the body
so it is given as



The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

<h3>Further explanation</h3>
Let's recall Impulse formula as follows:

<em>where:</em>
<em>I = impulse on the object ( kg m/s )</em>
<em>∑F = net force acting on object ( kg m /s² = Newton )</em>
<em>t = elapsed time ( s )</em>
Let us now tackle the problem!

<u>Given:</u>
mass of ball = m = 0.500 kg
initial speed of ball = vo = 20.0 m/s
final kinetic energy = Ek = 70% Eko
<u>Asked:</u>
magnitude of the change of momentum of the stone = Δp = ?
<u>Solution:</u>
<em>Firstly, we will calculate the final speed of the ball as follows:</em>



→ <em>negative sign due to ball rebounds</em>


<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

![\Delta p_{stone} = - [ mv - mv_o ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20-%20%5B%20mv%20-%20mv_o%20%5D)
![\Delta p_{stone} = m[ v_o - v ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20-%20v%20%5D)
![\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20m%5B%20v_o%20%2B%20v_o%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%20mv_o%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)
![\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]](https://tex.z-dn.net/?f=%5CDelta%20p_%7Bstone%7D%20%3D%200.500%20%28%2020.0%20%29%20%5B%201%20%2B%20%5Csqrt%7B0.7%7D%20%5D)


<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Dynamics
Answer:
Using the new cylinder the heat rate between the reservoirs would be 50 W
Explanation:
- Conduction could be described by the Law of Fourierin the form:
where
is the rate of heat transferred by conduction,
is the thermal conductivity of the material,
and
are the temperatures of each heat deposit,
is the cross area to the flow of heat, and
is the distance that the flow of heat has to go. - For the original cylinder the Fourier's law would be:
, and if
, then the expression would be:
where
is the diameter of the original cylinder, and
is the length of the original cylinder. - For the new cylinder, in the same fashion that for the first, Fourier's Law would be:
,where
is the heat rate in the second case,
and
are the new diameter and length. - But,
and
, substituting in the expression for
:
. - Rearranging:
. - In the last declaration of
, it could be noted that the expressión inside the parenthesis is actually
, then:
. - <u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>
Answer: 14.52*10^6 m/s
Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.
the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:
e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s