Question

MOMENTUM AND IMPULSE. I WILL GIVE MEDALSSSSS

Answer 1

Currently I am doing a take home test and since no one in my house knows a thing about Physics I would like one of you lovely people to check my answers before I submit my quiz. If interested just reply or shoot me a message.
It's okay to reply even if you see someone else in here because it's always okay to have your checked twice.

Answer 2

Would the answer to this be 90? Really the answer should be 91.3 but considering 91.3 is close to 90 I chose it.

Answer 3

Wait my answer is indeed 90 so you may disregard it or check my answer all together to ensure I am correct.

This problem here is confusing to me. I have no idea how to set up in order to solve. If I set it up the traditional way f(t)=J=m(delta)v I would need a mass and I was not given one in this particular problem

Answer 4

how did you get 90?

Answer 5

I set my equation as 83(0.0033)=(0.003)delta v
0.2739=0.003
This gave me an answer of 91.3

Then I chopped 0.2739 into 0.27,then divided by 0.003 and that yielded an answer of 90

Answer 6

don't chop it, and you get 91.3

Answer 7

But there is not answer choice for 91.3, only 90 and some other irrelevant answer choices are present

Answer 8

oh, it is multiple choice.

we get 91.3 m/s, and because it is final numerical result, now we round it to significant figures.

Of the numbers given in the question 0.003 kg is the least certain, it has only one significant figure.

Therefore our numbercal answer is only valid to one significant figure, i.e. 90 m/s

Answer 9

so you have the first one right.

For the second one, use J = F∆t

Answer 10

as F = ma = m ∆v / ∆t

so m = F ∆t / ∆v

i.e J = m ∆v
= F ∆t

Answer 11

If I used the equation you've suggested for question two I would get an answer of 114 which is not one of the answer choices

Answer 12

But I don't know my mass, I only have force an time so how would I find my mass from that?

Answer 13

hmmm, i'm getting 110 N s

Answer 14

How are the blocks being broken, is there a bounce or not?

Answer 15

It doesn't say

Answer 16

Wanna try a different problem?

Answer 17

100 seems like the best option

Answer 18

\[\qquad\quad\ \ p_\text{john} = p_\text{running back}\\[2ex]
m_\text{john}\times v_\text{john} = m_\text{running back}\times v_\text{running back}\]

Answer 19

\[v_\text{john} = \]

Answer 20

7.2!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 21

yep

Answer 22

I got 1.004 for the last problem, I posted

Answer 23

I'm getting 1.984 m

Answer 24

Just after the collision,we can apply conservation of linear momentum to get the initial velocity before the block+bullet slide over frictional surface :
\[0.004*500 + 1*0 = (0.004+1)*v\tag{1}\]

Answer 25

The answer is 1.99 but I just rounded it to two. So we know that velocity final will be 1.99 (2) Now we need to use impulse momentum theory to find the time it takes to stop.

pf-pi=2-2=0

If I have done this correctly then my time is zero correct?

Answer 26

I think we may simply use the kinematics equations

Answer 27

\(v^2 = 2as\)

here, \(a = \dfrac{F}{m} = \dfrac{10*m*\mu}{m} = 10*\mu = 10*0.1 = 1\)

Answer 28

\(1.992^2 = 2*1*s\)

you can solve \(s\)

Answer 29

S=1.984 which rounds to 2 correct?

Answer 30

looks good to me, have they asked you to round ?

Answer 31

No,but I am sure my teacher would want me to round. Hey do you mind checking my other answers before I submit, I can take a screenshot, that is if you don't mind doing it

Answer 32

please open a new post for each question
il try...

Answer 33

Okay!