A perfect elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed into another form of energy in the collision. Well hope this answers your question :)
Answer:
F = 1618.65[N]
Explanation:
To solve this problem we use the following equation that relates the mass, density and volume of the body to the floating force.
We know that the density of wood is equal to 750 [kg/m^3]
density = m / V
where:
m = mass = 165[kg]
V = volume [m^3]
V = m / density
V = 165 / 750
V = 0.22 [m^3]
The floating force is equal to:
F = density * g * V
F = 750*9.81*0.22
F = 1618.65[N]
If speed = distance/time , then time = speed/distance.
So...
Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)
Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))
I hope this was a helpful explanation, please reply if you have further questions about the problem.
Good luck!
Answer:
The mechanical energy of the helicopter is 
.
Explanation:
It is given that,
Mass of the helicopter, m = 3250 kg
Speed of the helicopter, v = 56.9 m/s
Position of the helicopter, h = 185 m
The energy possessed by an object due to its motion is called its kinetic energy. It is given by :
The energy possessed by an object due to its position is called its potential energy. It is given by :
The sum of kinetic and potential energy is called mechanical energy of the system. It is given by :
or
So, the mechanical energy of the helicopter is . Hence, this is the required solution.
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
