1. Harry is pushing a car down a level road at 2.0m/s with a force of 243N. The total force acting on the car in the opposite direction, including road friction and air resistance, is which of the following?

a) slightly more than 243N
b) exactly equal to 243N
c) slightly less than 243N

2. a 0.50kg model rocket accelerates from 20m/s [up] to 45 m/s [up] in 0.70s. Calculate the unbalanced force acting on it. (I got 18N. Is that correct?)

3. an 800km boat slows down uniformly from 50km/h [E] to 20km/h [E] as it enters the harbour. If the boat slows down over a 30m distance, what is the force of friction on the boat? (hint, convert velocities to m/s)

F = m a

if velocity is constant, a = 0
so
F = 0
Therefore
the force is exactly equal and opposite

2.
a = change in velocity/change in time
a = (45-20)/.7 = 35.71
F = m a = .5*35.71 = 17.85 N so yes

3. change in velocity = 20-50 = -30 km/hr
-30 km/hr * (1000 m/km)(1 hr/3600s)
= -8.33 m/s
so
a = -8.33/time
to find time
average speed = (50+20)/2 = 35 km/hr
35*(1000/3600) = 9.72 m/s
so
time deaccelerating = 30m/9.72m/s
= 3.09 seconds
so acceleration = -8.33/3.09 = -2.7 m/s^2
F = m a = 800 (-2.7) = -2159 Newtons

NOTE - this is not really right in the real world of boats. The boat is in water some of which partners in the acceleration of the boat creating what naval architects call "added mass" and a larger force would actually be required.

1. Well, if Harry is pushing a car down the road at 2.0m/s with a force of 243N, and there's friction and air resistance acting in the opposite direction, then the total force acting on the car must be...wait for it...sliiightly more than 243N. Yup, those pesky forces always know how to ruin a good push.

2. Ah, the model rocket question. Calculating the unbalanced force on it...let me get my scientific clown calculator out. *Honk honk* Yes, you got it right! The unbalanced force is indeed 18N. Good job, rocket scientist!

3. Ahoy, sailor! So, the boat is slowing down from 50km/h to 20km/h over a 30m distance. Let's convert those velocities to m/s to keep things consistent. *Takes out clown conversion chart* Okay, 50km/h is approximately 13.89m/s, and 20km/h is about 5.56m/s. Now, to find the force of friction on the boat, we need to calculate the acceleration first. Thankfully, we have a handy-dandy formula: acceleration = change in velocity / time. Since we know the change in velocity is (13.89 - 5.56)m/s and the time is...wait, what's the time? I feel like this problem is missing something. *Gives problem a suspicious look* Well, without the time, I'm afraid I can't calculate the force of friction for you. My clown physics powers only go so far, you know.

1. To find the total force acting on the car in the opposite direction, we need to consider the force Harry is applying and the opposing forces such as road friction and air resistance. Since Harry is pushing the car forward, the opposing forces will act in the opposite direction. According to Newton's second law of motion, the net force acting on an object equals the mass of the object multiplied by its acceleration. Therefore, we can calculate the total force as F = m*a, where F is the force, m is the mass of the car, and a is the acceleration.

Given that Harry is pushing the car at a constant velocity of 2.0 m/s, we can assume the car is not accelerating. This means that the net force acting on the car is zero, and the force Harry is exerting is canceled out by the opposing forces. Therefore, the total force acting on the car in the opposite direction is exactly equal to 243N. Hence, the correct answer is (b) exactly equal to 243N.

2. To calculate the unbalanced force acting on the model rocket, we can use Newton's second law of motion. The formula is F = m*a, where F is the force, m is the mass of the rocket, and a is the acceleration.

Given the initial velocity (u) of the rocket as 20 m/s, the final velocity (v) as 45 m/s, and the time (t) as 0.70 s, we can calculate the acceleration directly using the formula a = (v - u) / t.

a = (45 m/s - 20 m/s) / 0.70 s = 25 m/s / 0.70 s ≈ 35.71 m/s^2

Now, we can substitute the known values into the formula to find the unbalanced force:

F = m*a

F = 0.50 kg * 35.71 m/s^2

F ≈ 17.86 N

Therefore, the correct calculation for the unbalanced force acting on the model rocket is approximately 17.86 N. Your answer of 18N is close enough and can be considered correct.

3. To find the force of friction on the boat, we can use Newton's second law of motion. The formula is F = m*a, where F is the force, m is the mass of the boat, and a is the acceleration.

First, let's convert the velocities from kilometers per hour to meters per second.

Initial velocity = 50 km/h [E]
= (50 km/h) * (1000 m/km) / (3600 s/h)
= 13.89 m/s [E]

Final velocity = 20 km/h [E]
= (20 km/h) * (1000 m/km) / (3600 s/h)
= 5.56 m/s [E]

The boat slows down over a distance of 30 m. Since the boat slows down uniformly, we can use the equation of motion:

v^2 = u^2 + 2*a*s

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Plugging in the values, we can solve for the acceleration:

(5.56 m/s)^2 = (13.89 m/s)^2 + 2*a*(30 m)

30.86 m^2/s^2 - 192.78 m^2/s^2 = 2*a*(30 m)

-161.92 m^2/s^2 = 60*a

a ≈ -2.699 m/s^2

As the boat is slowing down, the acceleration is negative, indicating a retardation or deceleration.

Now, we can calculate the force of friction on the boat:

F = m*a

Given that the mass of the boat (m) is not mentioned in the question, we cannot determine the force of friction without this information.

To answer the questions, we need to apply relevant physics principles. Let's break down each question and explain how to arrive at the correct answer:

1. To determine the total force acting on the car, we need to consider the force Harry is exerting and the opposing forces such as friction and air resistance. Since the car is being pushed in the forward direction at a constant velocity, we can conclude that the total force acting on the car must be equal in magnitude and opposite in direction to the force Harry is exerting.

Therefore, the total force acting on the car in the opposite direction is b) exactly equal to 243N.

2. To calculate the unbalanced force acting on the model rocket, we can use the equation:

unbalanced force = (mass of the rocket) * (change in velocity) / (time interval)

In this case, the mass of the rocket is given as 0.50kg, the change in velocity is (45m/s - 20m/s) = 25m/s, and the time interval is 0.70s. Plugging these values into the equation:

unbalanced force = (0.50kg) * (25m/s) / (0.70s)
unbalanced force ≈ 17.86N

So, your answer of 18N is correct. Well done!

3. To calculate the force of friction on the boat, we can use Newton's second law of motion:

force = mass * acceleration

First, let's convert the given velocities from km/h to m/s. 50 km/h is equal to (50 * 1000) / 3600 = 13.89 m/s, and 20 km/h is equal to (20 * 1000) / 3600 = 5.56 m/s.

Next, we need to calculate the acceleration of the boat. Using the formula:

acceleration = (final velocity - initial velocity) / time

Here, the initial velocity is 13.89 m/s, the final velocity is 5.56 m/s, and the distance over which the boat slows down is 30m. We are not given the time, but we can find it using the equation:

time = distance / average velocity

average velocity = (initial velocity + final velocity) / 2

Substituting the values:

average velocity = (13.89 m/s + 5.56 m/s) / 2 = 9.72 m/s
time = 30m / 9.72 m/s ≈ 3.08s

Now, we can substitute the values into the equation for force:

force = (mass of the boat) * (acceleration)
force = (800,000g) * ((5.56 m/s - 13.89 m/s) / 3.08s)

Using the calculated values:

force ≈ 800,000g * (-2.83 m/s) / 3.08s

You'll need to provide the value of g (the acceleration due to gravity) to get the final numerical answer. However, the force of friction on the boat will be negative in this case since it opposes the motion and acts in the opposite direction of the boat's motion.

Hope this helps explain how to solve these problems!