The answer would be false
Answer:
No. 67
Peter Street
12th Road
Chennai
24th June 201_
Dear Amrish
I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.
It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?
I am anxiously waiting for your reply.
Yours affectionately
your name
Answer:
(5g/cm³)*(10cm³) = 50g
Explanation:
This is just a conversion formula. Easy to find using dimensional analysis.
(5g/cm³)*(10cm³) = 50g
According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>
In other words, this law states a relation between the orbital period
of a body (moon, planet, satellite) orbiting a greater body in space with the size
of its orbit.
This Law is originally expressed as follows:
<h2>

(1)
</h2>
Where;
is the Gravitational Constant and its value is 
is the mass of Jupiter
is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
<h2>

(2)
</h2>
Then:
<h2>

(3)
</h2>
Which is the same as:
<h2>

</h2>
Therefore, the answer is:
The orbital period of Io is 42.482 h
Answer:

Explanation:
Hello.
In this case, since the force is defined in terms of the mass and acceleration by:

We can easily compute the mass by solving for it:

Whereas the force is 182 N (kg*m/s²) and the acceleration is 13 m/s², therefore, we obtain:

Best regards.