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3 years ago

Define the term overload.

Physics

2 answers:

Burka [1]3 years ago

8 0

To load an excessive amount of something :)

ryzh [129]3 years ago

4 0

Answer:

Verb. Load with too great a burden or cargo.

"Both boats were overloaded and low in the water"

Noun. An excessive load or amount.

"an overload of stress"

Explanation:

Similar words are strain, excess, and overburden.

Have a good day and stay safe!

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What would most likely happen as a result of the generator in a wind turbine breaking? The blades would not be turned. Less stea

Aleonysh [2.5K]

Answer:

What would most likely happen as a result of the generator in a wind turbine breaking?

The What would most likely happen as a result of the generator in a wind turbine breaking?

The blades would not be turned.

Less steam would be produced.

Electricity would not be generated.

Solar energy would not be absorbed.

The blades would not be turned.

Less steam would be produced.

Electricity would not be generated.

Solar energy would not be absorbed.

Explanation:

6 0

3 years ago

Read 2 more answers

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

4 years ago

A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring balance C and D. Balance C is

vagabundo [1.1K]

A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively

<h3>What is gravity?</h3>

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.

As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end

The weight of the rod acting downward is from the center of the rod at 40 cm

Let us suppose the reading on the spring balance C and D are P and Q respectively

By using the equilibrium for the vertical force

Fv=0

P + C = 3.2

By using the equilibrium for the moment around the left corner

20×P+ 50×Q= 40 ×3.2

By solving for both P and Q from the above two equations we would get

P =1.06 and Q = 2.13

Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively

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brainly.com/question/4014727

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1 year ago

Two cylindrical resistors are made from same material and have the same length. When connected across the same battery, one diss

Ilia_Sergeevich [38]

Answer:

$\frac{d_2}{d_1} = \sqrt2 = 1.41$

Explanation:

As resistor is connected to the battery of constant EMF then the power across the resistor is given as

$P = \frac{E^2}{R}$

now if two resistors are made up of same material and of same length then due to different cross sectional area they both have different resistance

Due to different resistance they both will have different power

Since power is inversely depends on the resistance

So if the power is twice that of the other then the resistance must be half

so we have

$R_1 = \rho \frac{L}{A_1}$

$R_2 = \rho\frac{L}{A_2}$

since one resistance is half that of other resistance

So the area of one must be twice that of other

so we have

$\frac{A_2}{A_1} = 2$

$\frac{\pi d_2^2}{\pi d_1^2} = 2$

$d_2 = 1.41 d_1$

8 0

3 years ago

How can gravitational tugs from orbiting planets affect the motion of a star?.

gayaneshka [121]

Answer:

Gravitational tugs from orbiting planets don't affect the motion of a star. The star, being much larger than the planet, has a much smaller orbit. But it does move slightly. Explain how alien astronomers could deduce the existence of planets in our solar system by observing the Sun's motion.

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2 years ago