which statement best describes what would happen if the number of wore coils in a electromagnavneg were increased

Allisa [31]

Answer:

The SI unit of intensity is the watt per square meter/metre (W/m^2.)

Explanation:

Intensity is equal to the power transferred per unit area. Since power is measured in watts (W) and 1 W = 1 J/s, then intensity can be viewed as how fast energy goes through a certain area.

In physics, intensity is often used when studying light, sound, or other phenomena that involve waves or energy transfer. (With waves, the power value is taken as the average power transfer over the wave's period.)

8 0

2 years ago

at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g

Readme [11.4K]

The mass of water that must be raised is $5.25\cdot 10^7 kg$

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

$P_{out} = 0.70 P_{in}$

where $P_{in}$ is the power in input.

The power in input can be written as

$P_{in} = \frac{W}{t}$

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

$W=mgh$

where

m is the mass of water

$g=9.8 m/s^2$ is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

$P_{out} = 0.70 \frac{mgh}{t}$

Where

$P_{out} = 150 MW = 150\cdot 10^6 W$

And solving for m, we find:

$m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg$

Learn more about power:

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3 0

3 years ago

A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A

tino4ka555 [31]

Answer:

(a) Negative Q

(b) Positive Q

Explanation:

Charge is the inherent property of matter due to the transference of electrons.

There are three methods of charging a body.

(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.

(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.

(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.

(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.

(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.

7 0

2 years ago

A boy standing on a bridge above a river throws stone A vertically upward with an initial velocity =15m/s.2 seconds later he dro

Anastaziya [24]

I'm not sure if this is correct but it's what I'll do

This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.

Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )

Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12

Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2

4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t

Time for Stone B is 4s
Time for Stone A is 6s

7 0

3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$