The starting and ending points of the motion are the same.. . . . .
Answer/solution:
Given :
Mass =5kg
T 1 =20 C,T 2 =100 ∘C
ΔT=100−20=80 ∘C
Q=m×C×ΔT
where C= specific heat capacity of water
=4200J/(kgK)
Q=5×4200×80
=1680000 Joule.
=1680KJ
Answer:
Acceleration of proton will be ![a=0.67\times 10^{11}m/sec^2](https://tex.z-dn.net/?f=a%3D0.67%5Ctimes%2010%5E%7B11%7Dm%2Fsec%5E2)
Explanation:
We have given a proton is placed in an electric field of intensity of 700 N/C
So electric field E = 700 N/C
Mass of proton ![m=1.67\times 10^{-27}kg](https://tex.z-dn.net/?f=m%3D1.67%5Ctimes%2010%5E%7B-27%7Dkg)
Charge on proton ![e=1.6\times 10^{-19}C](https://tex.z-dn.net/?f=e%3D1.6%5Ctimes%2010%5E%7B-19%7DC)
So electric force on the proton ![F=qE=1.6\times 10^{-19}\times 700=1.120\times 10^{-16}N](https://tex.z-dn.net/?f=F%3DqE%3D1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20700%3D1.120%5Ctimes%2010%5E%7B-16%7DN)
This force will be equal to force due to acceleration of the proton
According to newton's law force is given by F = ma
So ![1.67\times 10^{-27}\times a=1.120\times 10^{-16}](https://tex.z-dn.net/?f=1.67%5Ctimes%2010%5E%7B-27%7D%5Ctimes%20a%3D1.120%5Ctimes%2010%5E%7B-16%7D)
![a=0.67\times 10^{11}m/sec^2](https://tex.z-dn.net/?f=a%3D0.67%5Ctimes%2010%5E%7B11%7Dm%2Fsec%5E2)
So acceleration of proton will be ![a=0.67\times 10^{11}m/sec^2](https://tex.z-dn.net/?f=a%3D0.67%5Ctimes%2010%5E%7B11%7Dm%2Fsec%5E2)
A) The acceleration is due to gravity at any given point if you look at it vertically, so
![-10 m/s^2](https://tex.z-dn.net/?f=-10%20m%2Fs%5E2)
.
b)
![sin(25) = V_y/V](https://tex.z-dn.net/?f=sin%2825%29%20%3D%20V_y%2FV)
, so
![V_y = V*sin(25)](https://tex.z-dn.net/?f=V_y%20%3D%20V%2Asin%2825%29)
. We use
![V = V_0 + a t](https://tex.z-dn.net/?f=V%20%3D%20V_0%20%2B%20a%20t)
and then the final speed must be 0 because it stops at the highest point. So
![0 = V_y - 10t](https://tex.z-dn.net/?f=0%20%3D%20V_y%20-%2010t)
. Solve for
![t](https://tex.z-dn.net/?f=t)
and you get
![t = 32sin(25)/10 = 16sin(25)/5](https://tex.z-dn.net/?f=t%20%3D%2032sin%2825%29%2F10%20%3D%2016sin%2825%29%2F5)
c)
![Y = Y_0 + V_0t + (1/2)at^2](https://tex.z-dn.net/?f=Y%20%3D%20Y_0%20%2B%20V_0t%20%2B%20%281%2F2%29at%5E2)
, and then we plug the values:
![Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%2032sin%2825%29%2At%20-%20%281%2F2%29%2A10%2At%5E2)
and we already have the time from "b)", so
![Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%20%5B%2832sin%2825%29%29%2A%2832sin%2825%29%2F10%29%5D%20-%205%2832sin%2825%29%2F10%29%5E2)
; then we just rearrange it
![Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%2010%5B%2832sin%2825%29%29%5E2%2F100%5D%20-%205%20%5B%2832sin%2825%29%29%5E2%2F100%5D%20)
and finally
Answer:
A 0.17 kg baseball is launched from the roof of a building 14 m above the ground. Its initial velocity is 29 m/s at 40° above the horizontal. Assume any effects of air resistance are negligible.
(a) What is the maximum height above the ground that the ball reaches?
m
(b) What is the speed of the ball as it strikes the ground?
m/s