F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
Answer:
Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.
Where the weight of the block has an x-component and y-component:
(1)
(2)
As well as the Normal Force :
(3)
(4)
In addition, we know , then
In the X-component:
(5)
Substituting (1) in (5):
(6)
In addition, we know , where
is the mass of the block and
the gravity acceleration, which is equal to
So:
(7)
(8)
(9) >>>>This is the acceleration of the block
On the other hand, we have the following equation that expresses a <u>relation between</u> the distance with the acceleration
and time
:
(10)
We already know the value of and calculated
, we have to find
:
(11)
(12)
(13) >>>This is the time it takes to the block to go from the initial velocity
to its final velocity
If the acceleration is the variation of the velocity in time, we can use the following equation to find :
(13)
If
(14)
(15)
Finally we get the value of the Final Velocity of the block:
