Explanation:
The gravitational force equation is the following:

Where:
G = Gravitational constant = 
m1 & m2 = the mass of two related objects
r = distance between the two related objects
The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.
Answer:
the period of the 16 m pendulum is twice the period of the 4 m pendulum
Explanation:
Recall that the period (T) of a pendulum of length (L) is defined as:

where "g" is the local acceleration of gravity.
SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.
Differentiation in its simplest of terms means breaking something into small parts. On the other hand, integration is taking those really small parts and gluing them in the right order. In short, these terms are the direct opposite or inverses of each other. The term which can tell you how fast you are going at a moment in time at ones current location is called a derivative. The term on the other hand, which can tell you how far you have travelled if you have been keeping track of your location and your time is what an integral is referred to. It is like differentiation only needs knowledge on the local neighbourhood while integration will need the knowledge on a global knowledge.
Answer:
a) 2.85 kW
b) $ 432
c) $ 76.95
Explanation:
Average price of electricity = 1 $/40 MJ
Q = 20 kW
Heat energy production = 20.0 KJ/s
Coefficient of performance, K = 7
also
K=(QH)/Win
Now,
Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7
where
P(in) is the input power
Thus,
P(in) = 20/7 = 2.85 kW
b) Cost = Energy consumed × charges
Cost = ($1/40000kWh) × (16kW × 300 × 3600s)
cost = $ 432
c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95
Given data:
- It is a graphical display where the data is grouped in to ranges
- A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.
- It is an accurate representation of the distribution of numerical data.
<em>From Figure:</em>
Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).
<em>So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.</em>