Physics questions!

- korosh23

Physics questions!

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- korosh23

##### 1 Attachment

- korosh23

Please answer the orange questions which there is a question mark. Each question in a different page. So three medals.

- korosh23

Ok #36 please! I have my answer. I will send it in a sec!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- korosh23

##### 1 Attachment

- Michele_Laino

I got this:
|dw:1433908300305:dw|
so I can write this:
\[\Large {m_1}{v_1} - {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v\]
from which I got:
v= +4 m/sec

- korosh23

How !?

- Michele_Laino

using my reference system, we can write:
\[\Large \begin{gathered}
{{\mathbf{v}}_{\mathbf{1}}} = \left( {{v_1},0} \right) \hfill \\
{{\mathbf{v}}_{\mathbf{2}}} = \left( { - {v_2},0} \right) \hfill \\
\end{gathered} \]

- korosh23

Are you sure that is
m1v1 - m2v2 or m1v1 + m2v2?

- Michele_Laino

no, since what is conserved is the vector which represents the total momentum, actually our equation is:
\[\Large {m_1}{{\mathbf{v}}_{\mathbf{1}}} + {m_2}{{\mathbf{v}}_{\mathbf{2}}} = \left( {{m_1} + {m_2}} \right){\mathbf{v}}\]

- korosh23

Exactly, it is a closed system. Energy is conserved.
Now lets plug in the numbers.

- Michele_Laino

now we have to go from vectors to components, and as I wrote before, we can write:
\[\begin{gathered}
{{\mathbf{v}}_{\mathbf{1}}} = \left( {{v_1},0} \right) \hfill \\
{{\mathbf{v}}_{\mathbf{2}}} = \left( { - {v_2},0} \right) \hfill \\
\end{gathered} \]

- korosh23

what is v2? Motor or truck?

- Michele_Laino

v_2 is the motor

- Michele_Laino

furthermore:
\[\Large {v_1} = \left| {{{\mathbf{v}}_{\mathbf{1}}}} \right|,\quad {v_2} = \left| {{{\mathbf{v}}_{\mathbf{2}}}} \right|\]

- korosh23

Wait a moment.

- korosh23

In the question it says the truck is going west. West is the negative direction. Motorist is going opposite of west which is east! It is postive velocity. Truck should have -ve velocity.

- Michele_Laino

ok! Then our correct drawing is:
|dw:1433908946956:dw|
and the previous equation will become:
\[\Large \begin{gathered}
- {m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v \hfill \\
{{\mathbf{v}}_{\mathbf{1}}} = \left( { - {v_1},0} \right) \hfill \\
{{\mathbf{v}}_{\mathbf{2}}} = \left( { + {v_2},0} \right) \hfill \\
\end{gathered} \]

- Michele_Laino

and the answer is:
v= -4 m/sec

- korosh23

That means I was right? I got -4 m/s

- Michele_Laino

yes!

- korosh23

Ok great, thank you!
I have to go, I will ask my other questions later.

- Michele_Laino

ok!

- korosh23

Just wondering are you here in openstudy tomorrow. On Thursday I have a physics exam. I was wondering if I have few quick question, Is it ok I ask you?

- Michele_Laino

yes! I will stay in OpenStudy tomorrow, and I can help you

- korosh23

At what time are you in openstudy, and when you leave?

- Michele_Laino

I will be here in OpenStudy at 6:00 am (Italy Time zone)

- korosh23

Ok I will manage mt time. Thank you, you are a very supportive tutor.

- Michele_Laino

:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.