23) an unfortunate bug splatters against the windshield of a moving car. compared to the force of the car on the bug, the force of the bug on the car is
a) smaller b) the same c) larger d) need more information to decide

24) another unfortunate bug splatters against the windshield of a moving car. compared to the deceleration of the car, the deceleration of the bug is
a) less b) the same c) greater d) need more information to decide

25) john is pulling a 40.0 kg cart with a horizontal force of 100 n. the force of friction of the cart is 20.0 n. the acceleration of the cart is
a) 2.00 m/s²
b) 2.50 m/s²
c) 20.0 m/s²
d) 200 m/s²

26) the force of friction experienced by a 76.5 kg cross - country skier sliding on unwaxed hickory skis along dry snow at a constant speed of 4.00 m/s is - 60.0 n. what is the coefficient of friction between unwaxed hickory skis and dry snow?
(diagram: a skier on a 30° incline, with forces ( f_n ) (upward normal force), ( f_g ) (downward gravitational force), ( f_f ) (leftward frictional force), ( f_a ) (rightward applied force) shown.)

Question

23) an unfortunate bug splatters against the windshield of a moving car. compared to the force of the car on the bug, the force of the bug on the car is
a) smaller  b) the same  c) larger  d) need more information to decide

24) another unfortunate bug splatters against the windshield of a moving car. compared to the deceleration of the car, the deceleration of the bug is
a) less  b) the same  c) greater  d) need more information to decide

25) john is pulling a 40.0 kg cart with a horizontal force of 100 n. the force of friction of the cart is 20.0 n. the acceleration of the cart is
a) 2.00 m/s²
b) 2.50 m/s²
c) 20.0 m/s²
d) 200 m/s²

26) the force of friction experienced by a 76.5 kg cross - country skier sliding on unwaxed hickory skis along dry snow at a constant speed of 4.00 m/s is - 60.0 n. what is the coefficient of friction between unwaxed hickory skis and dry snow?
(diagram: a skier on a 30° incline, with forces ( f_n ) (upward normal force), ( f_g ) (downward gravitational force), ( f_f ) (leftward frictional force), ( f_a ) (rightward applied force) shown.)

Accepted Answer

### Question 23
# Brief Explanations:
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. The force of the car on the bug (action) and the force of the bug on the car (reaction) are an action - reaction pair. So these two forces are equal in magnitude.
# Answer:
B) the same

### Question 24
# Brief Explanations:
We know from Newton's second law $F = ma$ (where $F$ is force, $m$ is mass, and $a$ is acceleration). The force on the bug and the force on the car are equal (from Newton's third law). The mass of the bug ($m_{bug}$) is much smaller than the mass of the car ($m_{car}$). Since $F = ma$, and $F_{bug}=F_{car}$, we can write $a=\frac{F}{m}$. For the bug, $a_{bug}=\frac{F}{m_{bug}}$ and for the car, $a_{car}=\frac{F}{m_{car}}$. Since $m_{bug}<m_{car}$, $\frac{F}{m_{bug}}>\frac{F}{m_{car}}$, so the deceleration of the bug is greater.
# Answer:
C) greater

### Question 25
# Explanation:
## Step 1: Identify the net force
The net force ($F_{net}$) acting on the cart is the applied force minus the frictional force. The applied force $F = 100\ N$ and the frictional force $f=20\ N$. So $F_{net}=F - f$.
$F_{net}=100\ N- 20\ N = 80\ N$
## Step 2: Use Newton's second law to find acceleration
Newton's second law is $F_{net}=ma$, where $m = 40.0\ kg$ (mass of the cart) and we need to find $a$ (acceleration). Rearranging the formula for $a$, we get $a=\frac{F_{net}}{m}$.
Substituting $F_{net} = 80\ N$ and $m = 40.0\ kg$ into the formula: $a=\frac{80\ N}{40.0\ kg}=2.00\ m/s^{2}$
# Answer:
A) $2.00\ m/s^{2}$

### Question 26
# Explanation:
## Step 1: Recall the formula for frictional force
The formula for the force of kinetic friction is $F_{f}=\mu F_{n}$, where $F_{f}$ is the frictional force, $\mu$ is the coefficient of kinetic friction, and $F_{n}$ is the normal force. When the skier is on a horizontal - like surface (the slope's normal force in the vertical direction, but since the skier is moving at constant speed, in the vertical direction, the normal force $F_{n}$ is equal to the weight of the skier $mg$ (assuming the slope's angle does not affect the normal force in the direction perpendicular to the slope, but in this case, since we are dealing with the force of friction and the skier's weight, and the skier is moving at constant speed, the normal force $F_{n}=mg$).
The mass of the skier $m = 76.5\ kg$, and $g = 9.8\ m/s^{2}$. So $F_{n}=mg=76.5\ kg\times9.8\ m/s^{2}=749.7\ N$
## Step 2: Solve for the coefficient of friction
We know that $F_{f}=\mu F_{n}$, and we want to find $\mu$. Rearranging the formula gives $\mu=\frac{F_{f}}{F_{n}}$. We are given that $F_{f}=- 60.0\ N$ (the negative sign indicates direction, we can use the magnitude for calculation, so $|F_{f}| = 60.0\ N$) and $F_{n}=749.7\ N$.
$\mu=\frac{60.0\ N}{749.7\ N}\approx0.0800$ (we can also calculate $F_{n}$ as $m\times g=76.5\times9.8 = 749.7$, and then $\mu=\frac{60}{749.7}\approx0.0800$)
# Answer:
The coefficient of friction is approximately $0.0800$ (if we use $g = 9.8\ m/s^{2}$). If we use $g = 9.81\ m/s^{2}$, $F_{n}=76.5\times9.81 = 750.465\ N$, and $\mu=\frac{60}{750.465}\approx0.0800$ as well.

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QUESTION IMAGE

Question

Explanation:

Question 23

Step 1: Identify the net force

Step 2: Use Newton's second law to find acceleration

Answer:

Question 24