<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:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.
Answer:
- <u>77.8 m/s, downward</u>
Explanation:
For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf
- Average speed = (Vf + Vi)/2
Also, by definition, the average speed is the distance divided by the time:
- Average speed = distance / time
Then:
Other kinematic equation for uniform acceleration is:
Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)
Replacing the known values we can set a system of two equations:
From (Vf + Vi)/2 = 300m/6.62s
(Vf + Vi) = 2 × 300m/6.62s
- Vf + Vi = 90.634 equation 1
From Vf = Vi + a×t
Vf - Vi = 9.8 (6.62)
- Vf - Vi = 64.876 equation 2
Adding the two equations:
- Vf = 77.8 m/s downward (velocities must be reported with their directions)
The relationship between the two is that air temperature changes the air pressure. For example, as the air warms up the molecules in the air become more active and they use up more individual space even though there is the same<span> number of molecules. This causes an </span>increase<span> in the air pressure.</span>
