# Physics I Midterm Exam in 2 days

• Asuncion
In summary, the car's average speed is 44 km/h while the average velocity is 73.1 km/h. It takes 1.63 hours for the car to travel this distance.
Asuncion

1. A car travels 95 km to the north at 70.0 km/h, then turns around and travels 21.9 km at 80.0 km/h.
What is the difference between the average speed and the average velocity on this trip? Answer 27 km/h.

2. How many nanoseconds does it take for a computer to perform one calculation if it performs
6.7 × 107 calculations/second? Answer 15 ns.

3. A baseball is hit with a bat and, as a result, its direction is completely reversed and its speed is
doubled. If the actual contact with the bat lasts 0.45 s, what is the ratio of the acceleration to the

4.A football kicker is attempting a field goal from 44 m out. The ball is kicked and just clears the
lower bar with a time of flight of 2.9 s. If the angle of the kick was 45°, what was the initial speed of

5. Find the orbital speed of an ice cube in the rings of Saturn, if the mass of Saturn is 5.67 × 1026 kg
and the rings have an average radius of 100,000 km. Answer 19.5 km/s

6. From what height off the surface of Earth should an object be dropped to initially experience an
acceleration of 0.5400 g? Answer 2298 km.

7. A 23 kg mass is connected to a nail on a frictionless table by a (massless) string of length 1.3 m. If
the tension in the string is 51 N while the mass moves in a uniform circle on the table, how long
does it take for the mass to make one complete revolution? Answer 4.8 s

Thanks so much for helping me. The answers came with the 50 question review, but these are the only questions I just can't seem to wrap my brain around.

Hi Asuncion!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

1. average speed = final displacement - initial displacement / time. In this case, car's displacement: 95-21.9=73.1 km. Then I divided 95/70 and 21.9/ 80. time = 1.63 hr. Then 73.1/1.63= 44. Wrong.

2. Nano = 10^-9. 6.7 * 10^7 is divided by 1, but not sure why? when you divide exponents you subtract the difference between the two. So exponent ends up to be 10^-2. I understand this is a dimensional analysis problem, but I must not be writing it down on paper correctly.

3. Not even sure how to begin this problem. Is it projectile problem? if so acceleration = -9.8. Initial velocity = 0 or vice verse…

4. Okay so 44 m / 2.9 s= 15.17 m/s then divide by sin 45. why is speed divided by sin 45. Is it because slope of tangent line.. Meaning 15.17 is cos (y axis) and 45sin is x-axis. tangent inverse = cos/sin…..sounds like I'm stretching a little too much…..

5. Orbital speed = sqrt g*r. First I had to get little g. Little g = big G times mass / radius^2. So, 6.67^-11*5.67 *10^26/ 1.0 *10 ^6 squared = 37818.9. Now, sort of 37818.9 * 1.0* 10^ 6 = 194470.8. wrong.

6. acceleration = velocity/ time. g= -9.8. Not sure how to begin.

7. Given
Mass = 23 kg
Length = 1.3 m
Force = 51 N
time = ?
Not sure which formula to use.

For number two, if you have $\frac{calculations}{second}$ how would you convert that to $\frac{seconds}{calculation}$?

2.[cal]\frac{}{}[/seconds]

Oops I was playing around with the format. Trying to get my fraction format like yours instead of using the *^/ symbols. gets confusing.

you want to put the things like "calculations" and "seconds" inside of the { } brackets

2. Here's how I wrote it down on paper: 1 cal , 1 ns / 10^-9 sec, 10 ^7 sec/ 6.7 cal = .15 cal / 10^-2. Everything cancels out expect for nanosecond. Then after subtracting the exponents I get 10^-2. Instead of dividing by .15 cal / 10 ^-2 I flipped the numerator and denominator. 10^-2/ .15 then just moved the decimal point to spaces to the right... Am I right

$\frac{calulations}{seconds}$

Aww HA! Thanks :!)

2. 1 cal $\frac{1 ns}{10^{9-}}$ $\frac^{7}{10}{6.7 cal}$

Okay so maybe I need more practice. :/

I'm a little confused about how you got "10 ^7 sec/ 6.7 cal = .15 cal / 10^-2"

are you saying $\frac{10^{7}sec}{6.7 cal}$ = $\frac{.15 cal}{10^{-2}}$?

because that doesn't make much sense

all you really need to do is take the reciprocal of $\frac{6.7*10^{7} calculations}{second}$ and then convert to nanoseconds if the question requires it

2. Here's how I set it up but I still come out with the wrong answer.

$\frac{1 cal}{10^{-9}}$ $\frac{1 sec}{6.7*10^{7}}$

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2. Sorry, if you haven't figured it out already I'm a Visual Learner. But Thank that you God, a Light bulb just went off in my head. $\frac{1 cal}{10^{-9} sec}$ $\frac{1 sec}{6.7 *10^{7}}$ When you add the exponents you get $\frac{1}{6.7*^10{-2}sec }$. YAY!

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the reason why you take the reciprocal is because that is just flipping the fraction upside down. Like the reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$, right? You take the reciprocal of something by dividing 1 by that number. Whenever I take the reciprocal I picture this on the paper:

where $\frac{A}{B}$ is the fraction that I am taking the reciprocal of

If A is calculations and B is seconds, then taking the reciprocal is going to give you seconds over calculations, or seconds per calculation.

Oh SHISHKABOB Your the best! I found my mistake I shouldn't have been separating 6.7 * $10^{7}$. thanks again SHISKABOB I was making it too complicated! I'm trying not to I always make the easy stuff complicated, but the complicated stuff comes easy... why me

I have to admit this one was tricky because normally (well the way I learned in chemistry) the given units come first then the desired conversion factor comes next. In this case the desired conversion factor came first (or as you say take the reciprocal) then the given. Well We knocked #2 out. Want to try anymore? lol

In number 1 it appears that you determined the correct average speed, but now how would you determine the average velocity? Since it's asking for the difference between the average speed and the average velocity.

EDIT: actually it seems you've determined the average velocity

average velocity is found by dividing the final displacement by the total time taken
average speed is found by dividing the total distance traveled by the total time taken

or

$velocity_{avg}$ = $\frac{final\,displacement}{total\,time\,taken}$

$speed_{avg}$ = $\frac{total\,distance\,traveled}{total\,time\,taken}$

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1. $speed_{avg}$ = 71 mph and $velocity_{avg}$ = 44 mph. Therefore, 71 mph -44 mph = 27 mph.

that's what I got too

for number 3, no it's not a projectile motion problem, it is just asking about the ratio between the velocity of the ball just before it hits the bat and the acceleration of the bat as it reverses direction and doubles its speed

so basically it wants to know what acceleration would be needed to result in double the speed in the opposite direction, and the acceleration only lasts for .45 seconds. Then compare that to the original velocity.EDIT: oh and in problem 1, it's in km/h, most teachers are real sticklers about units so remember to be careful about putting down the right one ;)

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3. I'm going to try. Going back to pre-calculus-solving exponential equations w/ same base. Let original velocity = u.
v=$\frac{-2u}{.45}$
a=$\frac{-3u}{.45}$

Then $\frac{-2}{-3}$ = .66666667 or 6.7$s^{-1}$ or 6.7*$10^{-1}$...

well, it's close, I think you have the right idea.

Try thinking about it this way. Before the impact, what is the velocity of the baseball?

And then, what is the acceleration of the ball from original velocity to final velocity? Remember that acceleration = the change in velocity divided by the time taken to change velocity.

or a = $\frac{Δv}{Δt}$

we already know Δt, so what is Δv?

here is a picture to help visualize the velocity before impact (red) and after impact (blue)

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also, the ratio that the problem wants is acceleration to velocity, which means a/v

Thanks SHISHKABOB! I have to take a break for an hr or two. Gotta get dinner ready for my husband and three kids, but I'll be back! ;)

Since you put it that way. Δv=v. However, the problem states that the speed doubled plus its going in the opposite direction, that's where I got the -2.

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-2v minus v = -v? Remember that Δv is (vfinal - vinitial)

## What topics will be covered on the Physics I Midterm Exam?

The exam will cover topics such as kinematics, forces, energy, momentum, and circular motion.

## How should I prepare for the Physics I Midterm Exam?

To prepare for the exam, review your class notes, complete practice problems, and go over any assigned homework or worksheets. It is also helpful to work through past exams or quizzes.

## How long will the Physics I Midterm Exam be?

The exam will be 2 hours long.

## Will there be any calculators or formulas provided for the Physics I Midterm Exam?

No, you will need to bring your own calculator and any necessary formulas will need to be memorized.

## What should I expect on the Physics I Midterm Exam?

The exam will consist of a combination of multiple choice, short answer, and calculation problems. It will test your understanding of key concepts and your ability to apply them to different scenarios.

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