A) We differentiate the expression for velocity to obtain an expression for acceleration:
v(t) = 1 - sin(2πt)
dv/dt = -2πcos(2πt)
a = -2πcos(2πt)
b) Any value of t can be plugged in as long as it is greater than or equal to 0.
c) we integrate the expression of velocity to find an expression for displacement:
∫v(t) dt = ∫ 1 - sin(2πt) dt
x(t) = t + cos(2πt)/2π + c
x(0) = 0
0 = = + cos(0)/2π + c
c = -1/2π
x(t) = t + cos(2πt)/2π -1/2π
Answer:
8.0 N
Explanation:
Force: This can be defined as the mass of a body and its acceleration. The S.I unit of Force is Newton (N).
Mathematically, Fore is expressed as
F = ma ........................... equation 1
Where F = force, m = mass, a = acceleration.
and
I = mΔv
Δv = I/m ............................ Equation 2
Where I = impulse, m = mass, Δv = change in velocity
Given: I = 6.0 Newton-seconds, m = 0.1 kilogram.
Substituting into equation 2
Δv = 6.0/0.1
Δv = 60 m/s.
But
a = Δv/t
where t = time = 0.75 seconds.
a = 60/0.75
a = 80 m/s²
Substitute the values of a and m into equation 1.
F = 0.1(80)
F = 8.0 N.
Thus the average force produced = 8.0 N
Answer:
to be an eginere u would have to go to college and study hard
Explanation:
I think you're saying that once you start pushing on the cars, you want to be able to stop each one in the same time.
This is sneaky. At first, I thought it must be both 'c' and 'd'. But it's not
kinetic energy, for reasons I'm not ambitious enough to go into.
(And besides, there's no great honor awarded around here for explaining
why any given choice is NOT the answer.)
The answer is momentum.
Momentum is (mass x speed). Change in momentum is (force x time).
No matter the weight (mass) or speed of the car, the one with the greater
momentum is always the one that will require the greater (force x time)
to stop it. If the time is the same for any car, then more momentum
will always require more force.
A. because as the merry-go-round spins the child accelerates towards the center of the merry-go-round at a uniform rate.