Answer:
3.90 degrees
Explanation:
Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is
W = mg = 30*9.81 = 294.3 N
This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.
We can calculate the parallel one since it's the one that affects the force required to push up
F = WsinΘ
Since customer would not complain if the force is no more than 20N
F = 20
So the ramp cannot be larger than 3.9 degrees
Answer:
Vr = 3.24m/s
The boat is going 3.24m/s relative to the bank of the river.
Explanation:
The relative speed of the boat to the bank Vr is the resultant of speed of boat relative to the water Vb and the speed of boat as a result of the water current or wind Vw
Vr = √(Vb^2 + Vw^2) .....1
Given;
Vb = 2.6m/s
Vw = distance downstream/time = 690m/355s
Vw = 1.94m/s
From equation 1 above; substituting the values
Vr = √(2.6^2 + 1.94^2)
Vr = 3.24m/s
The boat is going 3.24m/s relative to the bank of the river.
Well, the relationship between the net force and mass and acceleration of an object are directly related, as per the equation - Fnet = ma.
Thus the solution is A. As the net force of an object decreases, the object's acceleration also decreases, mass is kept constant.
Answer:
The acceleration of the crate is 
.
Explanation:
Given that,
Force, F = 750 N
Mass of the crate, m = 250 kg
The coefficient of friction is 0.12.
We need to find the acceleration of the crate. The net force acting on the crate is given by :
f is frictional force, 
So, the acceleration of the crate is . Hence, this is the required solution.
