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hodyreva [135]
3 years ago
12

Q. Calculate the annual energy

Physics
1 answer:
melomori [17]3 years ago
3 0

Answer:

60 kWh

Explanation:

The computation of the annual energy consumption in KW-h is shown below:

As we know that

1 kw = 1000 w

So, for 1400 W it would be

= 1,400 ÷ 1,000

= 1.4 kW

Now the number of hours it used in a year

= 7 minutes × 365 days ÷ 60 minutes

= 42.58333 hours

So in one year it used

= 1.4 kW × 42.58333

= 59.61 kWh

= 60 kWh

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Researchers at the University of Georgia have evaluated trends in streamside forests in areas within roughly 400 feet of the sta
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Vector a with arrow has a magnitude of 12.3 units and points due west. vector b with arrow points due north. (a) what is the mag
lesantik [10]

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:

|b|=\sqrt{(a+b)^2-a^2}=\sqrt{(14.5)^2-(12.3)^2}=7.68


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):

tan x=\frac{b}{a}=\frac{7.68}{12.3}=0.62

And the angle is

x=tan^{-1} (0.62)=31.8^{\circ}

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, 31.8^{\circ}, but this time the direction is south-west instead of north-west.

5 0
4 years ago
(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical
musickatia [10]

(a)

mass of the car: m=950 kg

Initial speed: u = 90.0 km/h=25 m/s

Final speed: v=0 (the car comes to rest)

distance: S=120 m

We can find the acceleration by using the following SUVAT equation:

v^2 -u^2 =2aS

Re-arranging it and replacing the numbers, we find the acceleration

a=\frac{v^2-u^2}{2S}=\frac{0-(25 m/s)^2}{2(120 m)}=-2.6 m/s^2

So now we can calculate the force using Newton's second law:

F=ma=(950 kg)(-2.6 m/s^2)=-2470 N

And the negative sign means the force is applied against the direction of motion.

(b)

In this case, the distance is different:

S=2.00 m

so, the acceleration in this case is

a=\frac{v^2-u^2}{2S}=\frac{0-(25 m/s)^2}{2(2 m)}=-156.3 m/s^2

And so, the force applied in this case is

F=ma=(950 kg)(-156.3 m/s^2)=-148 500 N

which is much larger than the force exerted in the previous exercise.


6 0
4 years ago
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