<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:
129 J/Kg°C
Explanation:
Given :
Mass of gold, m = 1.2kg
Quantity of heat applied, Q = 3096 J
Temperature, t2 = 40°C
Temperature, t1 = 20°C
Change in temperature, dt = (40-20)°C = 20°C
Using the relation :
Q = mcdt
Where, C = specific heat capacity of gold
3096 = 1.2kg * C * 20°C
3096 J = 24kg°C * C
C = 3096 J / 24 kg°C
C = 129 J/Kg°C
Answer:

Explanation:
Hello,
In this case, considering that the acceleration is computed as follows:

Whereas the final velocity is 28.82 m/s, the initial one is 0 m/s and the time is 4.2 s. Thus, the acceleration turns out:

Regards.
Answer:
c. The coefficient of kinetic friction is less than the coefficient of static friction
Explanation:
When the box finally does break loose. Then the component of the box weight which is parallel to the board weight parallel component, is equal to the
.

For the box to acce;erate thee must be non-zero net force acting on the box parallel to the board. Or we can say,

Therefore the force of kinetic friction must be less than the force of static friction. Thus,
