F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
Answer:
(a) 0.017m/s^2
(b) 17/100,000
(c) 0.17m, 0.558ft
Explanation:
(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s
Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2
(b) g = 9.8m/s^2, a = 0.0017m/s^2
a/g = 0.0017/9.8 = 0.00017 = 17/100,000
(c) Distance = speed × time = 0.017m/s × 10s = 0.17m
Distance in foot = 0.17 × 3.2808ft = 0.558ft
Answer:
Explanation:
Given ,
dv / dt = k ( 160 - v )
dv / ( 160 - v ) = kdt
ln ( 160 - v ) = kt + c , where c is a constant
when t = 0 , v = 0
Putting the values , we have
c = ln 160
ln ( 160 - v ) = kt + ln 160
ln ( 160 - v / 160 ) = kt
(160 - v ) / 160 = 
1 - v / 160 = 
v / 160 = 1 -
v = 160 ( 1 - )
differentiating ,
dv / dt = - 160k
acceleration a = - 160k
given when t = 0 , a = 280
280 = - 160 k
k = - 175
a = - 160 x - 175
a = 28000
when a = 128 t = ?
128 = 28000
= .00457
You're a little late. But if you want some short, quick rules, then these are
a couple that I would take in with me (stored only in my brain, of course):
-- If something is not accelerating or moving at all, then all the forces on it
must add up to zero. That could even mean a hanging rope.
-- In a vertical rope, the tension in it is the same everywhere in the rope.
The tension is the weight of whatever is hanging from the bottom.
That's really all I'm sure of, based on your hazy, fuzzy description of
what you've been doing in class. I don't want to get into things that
you might not have learned yet, and confuse you.
Bout 1000 ponies to the crank and she will get up and go
