Answer:
A) s = 796.38 m
B) t = 12.742 s
C) T = 25.484 s
Explanation:
A) First of all let's find the time it takes to get to maximum height using Newton's first equation of motion.
v = u + gt
u = 125 m/s
v = 0 m/s
g = 9.81 m/s²
Thus;
0 = 125 - 9.81(t)
g is negative because motion is against gravity. Thus;
9.81t = 125
t = 125/9.81
t = 12.742 s
Max height will be gotten from Newton's 2nd equation of motion;
s = ut + ½gt²
s = (125 × 12.742) + (½ × -9.81 × 12.742²)
s = 1592.75 - 796.37
s = 796.38 m
B) time to reach maximum height is;
t = u/g
t = 125/9.81
t = 12.742 s
C) Total time elapsed is;
T = 2u/g
T = 2 × 125/9.81
T = 25.484 s
-- The string is 1 m long. That's the radius of the circle that the mass is
traveling in. The circumference of the circle is (π) x (2R) = 2π meters .
-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .
-- Centripetal acceleration is V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²
-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =
64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .
</span>That's it. It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span>
</span>If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.
You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping.
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour ! This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.
Explanation:
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Answer:
40 N
Explanation:
We first need to calculate the acceleration of the tron ball.
Since acceleration, a = (v - u)/t where u = initial velocity of iron ball = 17m/s, v = final velocity of iron ball = 27m/s and t = time taken for the change in velocity = 5 s.
So, a = (v - u)/t
= (27 m/s - 17 m/s)/5 s
= 10 m/s ÷ 5 s
= 2 m/s²
We know force on iron ball, F = ma where m = mass of iron ball = 20 kg and a = acceleration = 2 m/s²
So, F = ma
= 20 kg × 2 m/s²
= 40 kgm/s²
= 40 N
So, the magnitude of the force on the iron ball is 40 N.
