Answer:
See answers below
Explanation:
a.
F = mg,
15.5 N = m(9.8 m/s²)
m = 1.58 kg
b.
Fnet = Applied force - resistance,
Fnet = 18 N - 4.30 N,
Fnet = 13.70 N
Fnet = ma
13.70 N = (1.58 kg)a
a = 8.67 m/s²
For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.
From the geometry of the problem, the 20 m-long cable creates
the hypotenuse of a right triangle, with the extended of the other two sides of
size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased
by 20 m - 17.3 m = 2.7 m.
The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 *
1.6 m , or about 37044 joules.
Answer:
(a) average velocity = 17.6 m/s
(b) when t = 0, v = 0
when t = 4, v = 19.2 m/s
when t = 8, v = 28.8 m/s
(c) after starting from rest, the car will be at rest again in 20 s
Explanation:
Given;
x(t)=bt²−ct³, substitute the given values and the equation will become;
x(t)=3t²−0.1t³
(a)average velocity = total distance / total time
total distance, x(t) = 3t²−0.1t³
x(8) = 3t²−0.1t³
X(8) = 3(8)² - 0.1(8)³
X(8) = 140.8 m
total time = 8 s
average velocity = 140.8 / 8
average velocity = 17.6 m/s
(b) instantaneous velocity = dx / dt
dx / dt = 6t - 0.3t²
when t = 0
v = 0
when t = 4 s
v = 6(4) - 0.3(4²) = 19.2 m/s
when t = 8 s
v = 6(8) - 0.3(8²) = 28.8 m/s
(c) the velocity is zero at dx / dt = 0
6t - 0.3t² = 0
t(6 - 0.3t) = 0
t = 0 or 6 - 0.3t = 0
t = 0 or 0.3t = 6
t = 0 or t = 6 / 0.3
t= 0 or t = 20 s
After starting from rest, the car will be at rest again in 20 s