Answer:
11.8 m/s
Explanation:
At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).
Sum of forces in the centripetal direction:
∑F = ma
mg − N = m v²/r
At the maximum speed, the normal force is 0.
mg = m v²/r
g = v²/r
v = √(gr)
v = √(9.8 m/s² × 14.2 m)
v = 11.8 m/s
Hi, the answer for this question is element!
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~Nate
W = force * displacement
W = 32 pounds * 10 feet
Now you need to convert it to newton and meters
W = 142 N * 3.048 m = 434 J
(I approximated the conversions- I hope it helps)
Velocity is the rate of change of position with respect to time, whereas acceleration is the rate of change of velocity. Both are vector quantities (and so also have a specified direction), but the units of velocity are meters per second while the units of acceleration are meters per second squared.
Answer:
total distance = 1868.478 m
Explanation:
given data
accelerate = 1.68 m/s²
time = 14.2 s
constant time = 68 s
speed = 3.70 m/s²
to find out
total distance
solution
we know train start at rest so final velocity will be after 14 .2 s is
velocity final = acceleration × time ..............1
final velocity = 1.68 × 14.2
final velocity = 23.856 m/s²
and for stop train we need time that is
final velocity = u + at
23.856 = 0 + 3.70(t)
t = 6.44 s
and
distance = ut + 1/2 × at² ...........2
here u is initial velocity and t is time for 14.2 sec
distance 1 = 0 + 1/2 × 1.68 (14.2)²
distance 1 = 169.37 m
and
distance for 68 sec
distance 2= final velocity × time
distance 2= 23.856 × 68
distance 2 = 1622.208 m
and
distance for 6.44 sec
distance 3 = ut + 1/2 × at²
distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²
distance 3 = 76.90 m
so
total distance = distance 1 + distance 2 + distance 3
total distance = 169.37 + 1622.208 + 76.90
total distance = 1868.478 m
