Answer:
No, The Moon, on the other hand, rotates once around its every 28 days, and once around the Earth in that same 28 days. The result of this combination is that the same side of the Moon is always facing the Earth.
Answer:
<em>The current is 11 Amperes</em>
Explanation:
<u>Electric Current</u>
The electric current is defined as a stream of charged particles that move through a conductive path.
The current intensity can be calculated as:
Where:
Q = Electric charge
t = Time taken by the charge to move through the conductor
The current intensity is often measured in Amperes.
The charge passing through a point in a circuit is Q= 55 c during t=5 seconds, thus the current intensity is:
I = 11 Amp
The current is 11 Amperes
1. electrical energy: electrical energy that is caused by moving electrons
2. coolant: a mixture of antifreeze and water that removes excess heat from an internal engine
3. electric compressor: a device that <span>acts as a pump, circulating refrigerant throughout the refrigerator
4. the inside of the fridge and the food becomes colder
5. the coolants becomes a hot, high-pressure gas
6. as coolant transfers thermal energy to the air outside, it turns back into a liquid
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Answer:
Explanation:
W = 75 watts
V = 110 volts
Formula
W = V * I
Solution
75 = 110 * I Divide by 110
75 / 110 = I
I = 0.6818 Amperes
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.
