Answer:
v = 2.94 m/s
Explanation:
Given that,
Mass of anvil, m = 15 kg
The gravitational potential energy relative to the floor, P = 65 J
We need to find the speed just before it hits the floor when the Anvil is released. Let it is v.
We can use the conservation of energy.
So, the required speed is 2.94 m/s.
part a)
Vector a has magnitude 12.3 and its direction is west, while Vector b has unknown magnitude and its direction is north. This means that the two vectors form a right-angle triangle, so a and b are two sides, while a+b is the hypothenuse.
We know the magnitude of a+b, which is 14.5, so we can use the Pythagorean theorem to calculate the magnitude of b:
part b) The direction of the vector a+b relative to west can be found by calculating the tangent of the angle of the right-angle triangle described in the previous part; the tangent of the angle is equal to the ratio between the opposite side (b) and the adjacent side (a):
And the angle is
with direction north-west.
part c)
This is exactly the same problem as the one we solved in part a): the only difference here is that the hypothenuse of the triangle is now given by a-b rather than a+b. In order to find a-b, we have to reverse the direction of b, which now points south. However, the calculations to get the magnitude of b are exactly the same as before, since the magnitude of (a-b) is the same as (a+b) (14.5 units), therefore the magnitude of b is still 7.68 units.
part d)
Again, this part is equivalent to part b); the only difference is that b points now south instead of north, so the vector (a-b) has direction south-west instead of north-west as before. Since the magnitude of the vectors involved are the same as part b), we still get the same angle, , but this time the direction is south-west instead of north-west.
B. Energy
A power company charges its customers for electricity based upon B. Energy.
<h3>
Explanation:</h3>
Kilo-watt Hours (kWh) is the unit that measures the electricity consumption of customers. Since Power is defined as the rate at which electrical energy is transferred by an electrical circuit per unit time,
If energy is transmitted at a constant rate over a period of time, the total energy in kilowatt hours is the product of power in kilowatts(kW) and time in hours (h)