Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²
Answer:
a= -0.86 m/s²
The negative sign shows that ball down the ground or moving down
Explanation:
Vf² - Vo² = 2gS
where
Vf = velocity of clay as it hits the ground
Vo = initial velocity of clay = 0
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
S = distance travelled by clay = 15 m
Substituting appropriate values,
Vf² - 0 = 2(9.8)(15)
Vf = 17.15 m/sec.
Formula to use is,
V - Vf = aT
where
V = velocity of clay when it stops = 0
Vf = 17.15 m/sec (as determined above)
a = acceleration
T = 20 ms
Put the values to find acceleration
a=(V-Vf)/T
a=(0-17.15)/20
a= -0.86 m/s²
The negative sign shows that ball down the ground
Well, it would be fossilization. <- If I spelled it correctly.
Answer:
Explanation:
We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:
where me(1) is mechanical energy while on h=10m
and me(2) is mechanical energy while on the ground
Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)
Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.
DynamicE(2) is equal to zero since it's touching the ground
Using that info we have
we divide both sides of the equation with mass to make the math easier.
The quantity work has to do with a force causing a displacement. Work has nothing to do with the amount of time that this force acts to cause the displacement. Sometimes, the work is done very quickly and other times the work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her body up a few meters along the side of a cliff. On the other hand, a trail hiker (who selects the easier path up the mountain) might elevate her body a few meters in a short amount of time. The two people might do the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The quantity that has to do with the rate at which a certain amount of work is done is known as the power. The hiker has a greater power rating than the rock climber.
Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.
Power = Work / time
or
P = W / t
