The acceleration exerted by the object of mass 10 kg is 
Answer: Option A
<u>Explanation:</u>
According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

So,

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

So, the acceleration exerted by the object of mass 10 kg is 
The magnitude of the vector C is 96.32m
<h3>How to solve for the magnitude of vector c</h3>
Ax = AcosθA
= 40 cOS 20
= 37.59
Ay = AsinθA
-40sin20
= -13.68
Bx = B cos θ B
= 75Cos50
= 48.21
By = BsinθB
= 75sin50
= 57.45
Cx = AX + Bx
= 37.59 + 48.21
= 85.8
Cy = Ay + By
= -13.65 + 57.45
= 43.77
The magnitude is solved by
|c| = 
= √85.8² + 43.77²
= 96.32m
The magnitude of the vector c is 96.32m
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Answer:
Explanation:
Given
Car speed decreases at a constant rate from 64 mi/h to 30 mi/h
in 3 sec


we know acceleration is given by 


negative indicates that it is stopping the car
Distance traveled



s=63.038 m
Answer: The bottom of the ladder is moving at 3.464ft/sec
Explanation:
The question defines a right angle triangle. Therefore using pythagorean
h^2 + l^2 = 10^2 = 100 ...eq1
dh/dt = -2ft/sec
dl/ dt = ?
Taking derivatives of time in eq 1 on both sides
2hdh/dt + 2ldl/dt = 0 ....eq2
Putting l = 5ft in eq2
h^ + 5^2 = 100
h^2 = 25 = 100
h Sqrt(75)
h = 8.66 ft
Put h = 8.66ft in eq2
2 × 8.66 × (-2) + 2 ×5 dl/dt
dl/dt = 17.32 / 5
dl/dt = 3.464ft/sec
The speed is 10.5 miles per minute, or 630 miles per hour.
We don't know the velocity, because you didn't tell us anything about the direction I'm going.