Question

If anyone could just help me with the formulas since my professor did not teach me anything that would be great! I think I am fine solving them.

Answer 1

@vocaloid

Answer 2

speed = distance/time

be careful with units. the time is given in minutes, but the answer choices ask for seconds.

Answer 3

Do I solve this the same?

Answer 4

a) average acceleration = (final velocity - initial velocity) / time

in this case, your final velocity is 0 (coming to a stop)
initial velocity is given as 33 km/hr
time is 19 minutes

make sure to convert all your units to meters and seconds

Answer 5

b) average velocity = (final speed - initial speed) / time

* important: normally, you need to consider the direction too, but in this case it's only asking for the magnitude

final speed is 0 since it's coming to a stop

initial speed is 33km/hr

time is 19 minutes

be sure to convert units to meters and seconds

Answer 6

I'm lost on a I got 29 and its wrong

Answer 7

33km/hr = 33*1000 m/hr = 33*1000/(3600) = 9.16666(repeated) m/s initial velocity

final velocity = 0

time = 19 minutes = 19*60 seconds = 1140 seconds

avg. acceleration = (0 - 9.16666)/1140 = -0.00804 m/s/s

Answer 8

It says that is wrong too

Answer 9

@smokeybrown I can't figure out what I'm doing wrong w/ my calculations (second problem, part a)

Answer 10

maybe take the negative out?

Answer 11

you have to take the negative out

Answer 12

so what do I do for b does a get used anywhere in the solving?

Answer 13

For part (b) of the question about the ship, I think the average velocity would be half of its initial velocity, assuming that the ship accelerates from max velocity to 0 at an equal rate.
So, unless I'm mistaken, that would be 33 km/hr divided by 2 (16.5 km/hr), only in meters per second instead of kilometers per hour.
16.5 km/hr = 16500 m/hr = 4.58 meters/second

I think that is the average velocity you are looking for

Answer 14

thank you! I would have solved it I just forgot the half part but I have two more questions that I may need help with

Answer 15

Sounds good. I'll be happy to take a look and try to help if I can

Answer 16

so how do I solve this

Answer 17

You found that the rate of acceleration is 0.64 m/s/s. So, the motorboat is gaining 0.64 m/s of velocity with every second that passes.
After 6.00 seconds, as the question mentions, the boat will gain 0.64*6.00 m/s of velocity, according to this acceleration. That equals an additional 3.84 m/s of velocity.
If you add this extra velocity to the velocity that the boat already has (9 m/s), you should be able to find the speed of the boat after 6 seconds have passed: 9 + 3.84 = ?

Answer 18

12.84

Answer 19

that makes sense

Answer 20

Yup, that should be the final velocity of the boat after accelerating at that rate for 6 more seconds.
Glad it makes sense :)

Answer 21

last one and then no more of this hw I will just study the content for my exam next week lol

Answer 22

This is supposed to be physical science for educators but then first day we are told that the content is not what we would teach upper elementary school students so yeah

Answer 23

Interesting. Sounds like a good plan. The question seems straightforward, so we can probably finish this up quick so you can go about your business.

We are given the initial height of the object and we're told that it is dropped from freefall; we can find out how long the object takes to reach the ground based on the equation:

\[Time = \sqrt{2*height/gravity}\]
Now, the gravity constant on earth is about 9.8, and the question says that the gravity constant on Uranus is about 0.889 times that on earth (0.889*9.8=0.871)

So, the amount of time the object takes to reach the ground on Earth would be:
\[Time = \sqrt{2*69/9.8}\]
= 3.75 seconds

And the amount of time the object takes to reach the ground on Uranus would be:
\[Time = \sqrt{2*69/8.71}\]
= 3.98 seconds

You can calculate the difference in the time it takes, in order to answer the question

Answer 24

.23 this is not that bad

Answer 25

thank you so much!!!

Answer 26

No problem! Glad we could help :)