Answer:
a. If an object's speed is constant, then its acceleration must be zero.
FALSE
As we know that acceleration is defined as the rate of change in velocity

so we can not say anything about the acceleration when speed is given to as and no information is given about velocity
b. If an object's acceleration is zero, then its speed must be constant.
TRUE
As we know that acceleration is defined as the rate of change in velocity

Since we know that if acceleration is 0 then velocity must be constant and hence speed is also constant
c. If an object's velocity is constant, then its speed must be constant.
TRUE
Since velocity is constant then it shows that its magnitude and direction both are constant so its speed is also constant.
d. If an object's acceleration is zero, its velocity must be constant.
TRUE
As we know that acceleration is defined as the rate of change in velocity

Since we know that if acceleration is 0 then velocity must be constant
e. If an object's speed is constant, then its velocity must be constant.
FALSE
Speed is just the magnitude so we can not say about its direction and hence if speed is constant then velocity may or may not change
Answer:
the velocity of the car is 0.875 m/s
Explanation:

therefore the V of car is 0.875 m
Answer:
The velocity of the student has after throwing the book is 0.0345 m/s.
Explanation:
Given that,
Mass of book =1.25 kg
Combined mass = 112 kg
Velocity of book = 3.61 m/s
Angle = 31°
We need to calculate the magnitude of the velocity of the student has after throwing the book
Using conservation of momentum along horizontal direction


Put the value into the formula


Hence, The velocity of the student has after throwing the book is 0.0345 m/s.
Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²