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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
Answer:
i believe the answer to you question is gravity and tension
Answer:
You are given that the mass of the clock M is 95 kg.
This is true whether the clock is in motion or not.
Fs is the frictional force required to keep the clock from moving.
Thus Fk = uk W = uk M g the force required to move clock at constant speed. (the kinetic frictional force)
uk = 560 N / 931 N = .644 since the weight of the clock is 931 N (95 * 9.8)
us is the frictional force requited to start the clock moving
us = static frictional force = 650 / 931 -= .698