10. the amount of net force required to keep a 5 - kg object moving rightward with a constant velocity of 2 m/s is ___.
a. 0 n b. 0.4 n c. 2 n d. 2.5 n e. 5 n
11. true or false:
for an object resting upon a non - accelerating surface, the normal force is equal to the weight of the object.
12. which one(s) of the following force diagrams depict an object moving to the right with a constant speed? list all that apply.
13. according to newtons third law, every force is accompanied by an equal and opposite reaction force. the reason that these forces do not cancel each other is ___.
a. the action force acts for a longer time period
b. the two forces are not always in the same direction
c. one of the two forces is greater than the other
d. the two forces act upon different objects; only forces on the same object can balance each other.
e.... nonsense! they do cancel each other. objects accelerate because of the presence of a third force.
15. a golf pro places a ball at rest on the tee, lines up his shot, draws back his club, and lets one rip. during the contact of the golf club with the golf ball, the force of the club on the ball is ___ the force of the ball on the club and the acceleration of the club is ___ than the acceleration of the ball.
a. greater than, greater than b. greater than, equal to c. greater than, less than
d. less than, less than e. less than, equal to f. less than, greater than
g. equal to, equal to h. equal to, greater than i. equal to, less than

Question

Accepted Answer

### 10.
# Explanation:
## Step1: Apply Newton's first law
According to Newton's first law of motion, an object moving with a constant velocity has zero acceleration ($a = 0$).
## Step2: Use Newton's second law
Newton's second law is $F_{net}=ma$. Since $a = 0$ and $m = 5$ kg, then $F_{net}=5	imes0=0$ N.

# Answer:
a. 0 N

### 11.
# Brief Explanations:
When an object is resting on a non - accelerating surface, the vertical forces are balanced. The normal force ($F_N$) acts upward and the weight ($W=mg$) acts downward. By Newton's first law (since there is no vertical acceleration, $a_y = 0$), $F_{net,y}=F_N - W=0$, so $F_N = W$.

# Answer:
TRUE

### 12.
# Brief Explanations:
For an object moving at a constant speed, the net force is zero. In force diagram a:
- Horizontal forces: $F_{left}=20$ N, $F_{right}=20$ N, so $F_{net,x}=20 - 20=0$ N
- Vertical forces: $F_{up}=25$ N, $F_{down}=25$ N, so $F_{net,y}=25 - 25=0$ N
In force diagram c:
- Horizontal forces: There are no horizontal forces (assuming no other horizontal forces not shown, but if we consider only the forces in the diagram,  $F_{net,x}=0$)
- Vertical forces: $F_{up}=25$ N, $F_{down}=25$ N, so $F_{net,y}=25 - 25=0$ N

In force diagram b:
- Horizontal forces: $F_{left}=15$ N, $F_{right}=20$ N, $F_{net,x}=20 - 15=5$ N (non - zero)
In force diagram d:
- Horizontal forces: Only $F_{right}=20$ N (non - zero net horizontal force)

# Answer:
a, c

### 13.
# Brief Explanations:
Newton's third - law force pairs (action - reaction) act on different objects. For forces to cancel each other (be in equilibrium), they must act on the same object.

# Answer:
d. the two forces act upon different objects; only forces on the same object can balance each other.

### 15.
# Brief Explanations:
According to Newton's third law, the force of the club on the ball and the force of the ball on the club are an action - reaction pair. So, $F_{club - on - ball}=F_{ball - on - club}$.
Using Newton's second law $F = ma$. Let $F$ be the magnitude of the force (same for both due to Newton's third law). If $m_{club}>m_{ball}$ (mass of the club is greater than the mass of the ball), then $a=\frac{F}{m}$. So, $a_{club}=\frac{F}{m_{club}}$ and $a_{ball}=\frac{F}{m_{ball}}$. Since $m_{club}>m_{ball}$, $a_{club}<a_{ball}$

# Answer:
i. equal to, less than

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QUESTION IMAGE

Question

Explanation:

10.

Step1: Apply Newton's first law

Step2: Use Newton's second law

Answer:

11.