The heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).
or C. Mass if you're on plato
Answer:
400ft. 32ft/s -32ft/s
Explanation:
In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2
Anyway for the sake of assumtion let us takes=160t-16t^2
ds/dt=160-32t=0
t=160/32= 5 seconds.
s=160*160/32-16*(160/32)^2= 400 mts
s=384 mts
160t-16t^2=384
i.e
16t^2-160t+384=0
t^2-10t+24=0
(t-6)(t-4)=0
t=[4,6]
we have to take t=4 because it is all the up i.e <5
velocity =v=ds/dt=160-32t
v=160-32*4=32 ft/sec still going up
for all the way down take t=6 whuch is >5
v=160-6*32=-32 ft/sec (falling down!!!)
Answer:
D
Explanation:
The decrease in potential energy is equal to the increase in kinetic energy.
mgh
250 x 9.8 x 30
=73, 500
The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha
Answer
given,
ω₁ = 0 rev/s
ω₂ = 6 rev/s
t = 11 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
11 α = 6 - 0
= 0.545 rev/s²
The angular displacement
θ₁= ωi t + (1/2) α t²
θ₁= 0 + (1/2) (0.545)(11)^2
θ₁= 33 rev
case 2
ω₁ = 6 rev/s
ω₂ = 0 rev/s
t = 14 s
Using equation of rotational motion
The angular acceleration is
ωf - ωi = α t
14 α = 0 - 6
= - 0.428 rev/s²
The angular displacement
θ₂= ωi t + (1/2) α t²
θ₂= 6 x 14 + (1/2) (-0.428)(14)^2
θ₂= 42 rev
total revolution in 25 s is equal to
θ = θ₁ + θ₂
θ = 33 + 42
θ = 75 rev
