Explanation:
Average acceleration is the change in velocity over change in time.
a = Δv / Δt
a = (63 m/s − 25 m/s) / (17 s − 6 s)
a ≈ 3.45 m/s²
Answer:
The speed of the car at the end of the 2nd second = 8.0 m/s
Explanation:
The equations of motion will be used to solve this problem.
A car starts from rest,
u = initial velocity of the car = 0 m/s
Accelerates at a constant rate in a straight line,
a = constant acceleration of the car = ?
In the first second the car moves a distance of 2.0 meters,
t = 1.0 s
x = distance covered = 2.0 m
x = ut + (1/2)at²
2 = 0 + (1/2)(a)(1²)
a = 4.0 m/s²
How fast will the car be moving at the end of the second second
Now,
a = 4.0 m/s²
u = initial velocity of the car at 0 seconds = 0 m/s
v = final velocity of the car at the end of the 2nd second = ?
t = 2.0 s
v = u + at
v = 0 + (4×2)
v = 8.0 m/s
Answer:
α=9.42 rad/s T= 0.014 rad/s
Explanation:
Low speed= Wi=3 rps = 3*2*π rad/s= 6π rad/sec
Medium speed= Wf= 9 rps= 9*2*π rad/sec= 18 rad/sec
time= t= 4 sec
Angular acceleration=α=
⇒α=(18 π - 6 π)/4
⇒α=9.42 rad/s
b).
Rotational inertia= I = 1.5*10^-3 kg-m^2
Net torque= T = I*α = (1.5*10^-3) * (9.42) rad/s^2
⇒T= 0.014 rad/s^2
If two variables are inversely proportional, then when one increases, the other decreases, and vice versa. If a variable, y, is inversely proportional to a variable, x, then y = k/x, where k is the proportionality constant.
Answer:
"Offgassing"
Explanation:
According to my research on Kinesiology, I can say that based on the information provided within the question the process being described is known as "Offgassing". In other words this process is defined as when something gives off or releases a chemical, especially a harmful one, in the form of a gas into the air..
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