True or false energy is always lost in a circuit

Physics

1 answer:

Llana [10]2 years ago

7 0

Explanation:

Electrons will gain energy as they are "pushed" from different points in the circuit. This energy is then lost when the electrons flow through circuit components such as a light bulb.

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If a marble is released from a height of 10 meters how long would it take to hit the ground?

zlopas [31]

S = ut + 0.5at^2

10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414} 

t^2 ~ 2.04 

t ~ 1.43 seconds

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3 years ago

show the trajectory of a body projected with an initial velocity vat an angle of departure thita is a paraboler

Jobisdone [24]

Answer:

Let me look up a couple of things regarding this question.

Explanation:

Then I will get back to you.

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2 years ago

“Anchor” points on a temperature scale are known as fiducial points. a)True b)False

Svetlanka [38]

Yes. It refers to any of the temperatures assigned to a number of reproducible equilibrium states on the International Practical Temperature Scale

In short, Your Answer would be "True"

Hope this helps!

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3 years ago

Read 2 more answers

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$