Answer:
In the absence of air resistance. I think no. D ) The bowling ball.
<em><u>Hope</u></em><em><u> this</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
<span>3.92 m/s^2
Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so
0.4 * 9.8 m/s^2 = 3.92 m/s^2
If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so
20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N
Multiply by the coefficient of static friction, giving
196 N * 0.4 = 78.4 N
So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass
78.4 N / 20.0 kg
= 78.4 kg*m/s^2 / 20.0 kg
= 3.92 m/s^2
And you get the same result.</span>
Answer:
The answer to your question is vo = 5.43 m/s
Explanation:
Data
distance = d= 5.8 m
height = 3 m
height 2 = 1.7 m
angle = 60°
vo = ?
g = 9.81 m/s²
Formula
hmax = vo²sinФ/ 2g
Solve for vo²
vo² = 2ghmax / sinФ
Substitution
vo² = 2(9.81)(3 - 1.7) / 0.866
Simplification
vo² = 19.62(1.3) / 0.866
vo² = 25.51 / 0.866
vo² = 29.45
Result
vo = 5.43 m/s
Answer:
1 second later the vehicle's velocity will be:

5 seconds later the vehicle's velocity will be:

Explanation:
Recall the formula for the velocity of an object under constant accelerated motion (with acceleration "
"):

Therefore, in this case
and 
so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

Objects are known to go down because of a unbalanced force