The energy possessed by a body because of its motion, equal to one half the mass of the body times the square of its speed is called its kinetic energy.
K=
2
1
mv
2
When, v=2v,
K
′
=
2
1
m(2v)
2
K
′
=4×
2
1
mv
2
K
′
=4K
Hence, when velocity is doubled, kinetic energy becomes 4 times.
Answer:
Total pressure exerted at bottom = 119785.71 N/m^2
Explanation:
given data:
volume of water in bottle = 150 L = 0.35 m^3
Area of bottle = 2 ft^2
density of water = 1000 kg/m
Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water
Pressure due to water P = F/A
F, force exerted by water = mg
m, mass of water = density * volume
= 1000*0.350 = 350 kg
F = 350*9.8 = 3430 N
A = 2 ft^2 = 0.1858 m^2
so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2
Atmospheric pressure
At sea level atmospheric pressure is 101325 Pa
Total pressure exerted at bottom = 18460.71 + 101325 = 119785.71 N/m^2
Total pressure exerted at bottom = 119785.71 N/m^2
General formula for emf is : emf = vBL (sin 0)...(1)
As the angle here is 90º and sin90º =1.
So, equation (1) becomes : emf = vBL Putting values :
emf = (6.2)(1.5)(3.96 * 10<span>^-3) = 0.0369 volts
</span>Hope this helps:)
Answer:
This is an example of Inelastic colission
Explanation:
Step one:
given:
mass of moose m1 = 620 kg
mass of train m2= 10,000kg
Initial velocity of moose u1= 0 m/s
Initial velocity of train v1 = 10m/s
combined velocity of the system is given as v
Applying the conservation of momentum equation we have
m1u1+ m2u1= (m1+m2)V
substitutting we have
620*0+10000*10= (620+10000)V
100000= 10620V
divide both sides by 10620
V = 100000/10620
V=9.41m/s
The velocity of the moose after impact is 9.41m/s
