When you are talking about the Principle of mechanical Energy Conservation, it is really only including the kinetic and potential energy in a total system. When frictional forces are present, although the conservation of energy law is still present, it does not work when it comes to the conservation of mechanical energy as there is another type of energy that is factored in. As friction acts on the object, that transition from potential to kinetic as it slide/falls will be completely different as some of that energy is being transformed into thermal energy. Which breaks the conservation of mechanical energy.
Answer it's 1,445 A
Explanation:
Basically 9,875 - what will give you 8,430 that's what's its asking. Its just a gap.
Answer:
The recoil velocity of the projectile is 0.0025m/s
Explanation:
Given:
Mass of the projectile =0.005kg
Mass of the launcher = 1500kg
Velocity = 750 m/s.
To Find:
The recoil velocity of the projectile = ?
Solution:
The recoil velocity is the obtained by dividing the "recoil momentum" by the "mass of the recoil body". The recoil momentum is equal to the momentum of the other body.
The momentum of the other body is equal to it mass times its velocity.
Lets find the recoil momentum,
Recoil momentum = mass of the projectile X velocity
Recoil momentum =
Recoil momentum = 3.75
Now Recoil Velocity,
Recoil Velocity = 
Recoil Velocity = 
Recoil Velocity = 0.0025m/s
<span>muscles perform Potential->Kinetic->heat potential- energy that can be used and is stored for use kinetic - movement heat- a prime factor of any kinetic relative.
hope it helped</span>
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>
