Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
![W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7BL%7D%20Fdx%20%3D%20%5Cint_%7B0%7D%5E%7BL%7Dax%20%5C%2C%20dx%20%3D%20%5Cfrac%7Ba%7D%7B2%7D%20%20%5Bx%5E%7B2%7D%20%5D_%7B0%7D%5E%7BL%7D%20%3D%20%20%5Cfrac%7BaL%5E%7B2%7D%7D%7B2%7D%20)
Answer:
Answer:

Explanation:
Let at any moment of time the friend's car is at some horizontal distance "x" from the position of balloon.
Now if the altitude of the balloon is fixed and it is at height "h"
so here we will have

now we know that
initially the angle of the friend's car is 35 degree
so the horizontal distance will be

similarly if the angle after passing the car position is 36 degree
then we have

now the speed of the balloon is constant
so we have




so the final position of friend when the angle is 36 degree


The leading theory of the driving force behind tectonic plate motion envisaged large scale convection current in the upper mantle, which can be transmitted throug the asthenosphere.
A :-) F = ma
Given - m = 95 kg
a = 2.2 m/s^2
Solution -
F = ma
F = 95 x 2.2
F = 209
.:. The force is 209 N
Answer:
V = 1.1 m/s
Explanation:
given,
mass of railroad car 1 , m = 30,000. kg
travelling at the speed , u = 2.2 m/s
mass of car 2, M = 30,000. kg
initial speed, u' = 0 m/s
final speed of the car after collision, V = ?
using conservation of momentum
m u + M u' = (M+m)V
30000 x 2.2 + 0 = (30000 + 30000) V
60000 V = 66000
V = 1.1 m/s
he velocity of the two cars is equal to V = 1.1 m/s