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ycow [4]
4 years ago
7

Can the answer to density be simplified​

Physics
1 answer:
kenny6666 [7]4 years ago
7 0

Answer: yes because all densities can be simplifed it just depends on the number.

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Technician A says that the battery must be in good condition and at least 75% charged to accurately test an alternator. Technici
aalyn [17]

Answer:Technician A

Explanation:

Technician A statement is correct as  

The battery is required to start the vehicle which, in effect, rotates the alternator at sufficient speed to keep the battery charged. This means if the battery is low it is not possible to start the vehicle and thus we are unable to test the alternator.  

That is the battery is pre-requisite to test the alternator. So the battery must be at least a 75 % charge to test the alternator.

5 0
3 years ago
A positively charged particle moves through an electric field. As part of a complicated trajectory, the particle passes through
kow [346]

Answer:

(B) The speed is larger at A than at B.

Explanation:

Point B, the final point of the trajectory, has higher electric potential than point A, the initial point of the trajectory, so the electric potential energy of the charged particle increases, which means that its kinetic energy must be decreasing, thus the speed at B must be lower than the speed at A.

8 0
3 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
A photon has 3.4 × 10–18 joules of energy. Planck’s constant is 6.63 × 10–34 J•s.
hodyreva [135]
Ok i will answer for real this time. Please give me brainliest.
<span>The Answerr is:
5.12*10^15. Since e=h*f, f=e/h. 3.4*10^(-18)/h.
</span>i am so sorry i was doing a challenge and i needed answers to get 100 pts.
Hope I Helped
~TeenOlafLover <3
5 0
3 years ago
the winner of a 10km road race took half an hour to complete the race. Calculate the average speed. Give your answer in metres p
leonid [27]
Hello there!

For this:

1). Convert 10km to meters!
2). Convert the 30 minutes into seconds!

3). Use the following formula to solve for speed!

speed= distance/time

Note: The units should automatically work out to m/s. :)

My goal is to make sure you understand the problem, which is why I won't be giving you the answer. It'll be more work now, but less work in the future! :)

Hope this helped!

-------------


DISCLAIMER: I am not a professional tutor or have any professional background in your subject. Please do not copy my work down, as that will only make things harder for you in the long run. Take the time to really understand this, and it'll make future problems easier. I am human, and may make mistakes, despite my best efforts. Again, I possess no professional background in your subject, so anything you do with my help will be your responsibility. Thank you for reading this, and have a wonderful day/night!
4 0
3 years ago
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