Answer:
Explanation:
An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:
(1)
where
is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, 
the angle of the slope
is the frictional force, with 
being the coefficient of friction and R the normal reaction of the incline
The equation of the forces along the direction perpendicular to the slope is
where
R is the normal reaction
is the component of the weight perpendicular to the slope
Solving for R,
And substituting into (1)
Re-arranging the equation,
This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.
Answer:
61578948 m/s
Explanation:
λ
= λ
687 = 570 
= 61578948 m/s
So Slick Willy was travelling at a speed of 61578948 m/s to observe this.
Answer:
The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid
B. The contact area of each tire with the track.
C. The co-efficent of static friction between the tires and the track.
D. The co-efficent of static friction between the tires and the track.
Explanation:
Answer:
i(t) = (E/R)[1 - exp(-Rt/L)]
Explanation:
E−vR−vL=0
E− iR− Ldi/dt = 0
E− iR = Ldi/dt
Separating te variables,
dt/L = di/(E - iR)
Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have
dt/L = -dx/Rx
-Rdt/L = dx/x
interating both sides, we have
∫-Rdt/L = ∫dx/x
-Rt/L + C = ㏑x
x = exp(-Rt/L + C)
x = exp(-Rt/L)exp(C) A = exp(C) we have
x = Aexp(-Rt/L) Substituting x = E - iR we have
E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So
E - i(0)R = Aexp(-R×0/L)
E - 0 = Aexp(0) = A × 1
E = A
So,
E - i(t)R = Eexp(-Rt/L)
i(t)R = E - Eexp(-Rt/L)
i(t)R = E(1 - exp(-Rt/L))
i(t) = (E/R)(1 - exp(-Rt/L))
Answer:
37357 sec
or 622 min
or 10.4 hrs
Explanation:
GIVEN DATA:
Lifting weight 80 kg
1 cal = 4184 J
from information given in question we have
one lb fat consist of 3500 calories = 3500 x 4184 J
= 14.644 x 10^6 J
Energy burns in 1 lift = m g h
= 80 x 9.8 x 1 = 784 J
lifts required 
= 18679
from the question,
1 lift in 2 sec.
so, total time = 18679 x 2 = 37357 sec
or 622 min
or 10.4 hrs
