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3 years ago

As they move through a DC circuit, electrons collide with ________ in resistors.

Physics

1 answer:

KengaRu [80]3 years ago

6 0

Answer:

idk man

Explanation:

i am suffereing from this i am in Mr.GoofFreinds class and this quiz is hard.

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A 75.0-kg ice skater moving at 10.0 m/s crashes into a stationary skater of equal mass. After the collision, the two skaters mov

Ksenya-84 [330]

Answer:

Explanation:

Momentum change for either skater is mΔv = 75.0(5.0) = 375 kg•m/s

As a change in momentum is equal to an impulse

375 = FΔt

F = 375/0.100 = 3750 N

As 3750 N < 4500 N no bones are broken.

4 0

2 years ago

A light ray moving from oil (n=1.38) into an unknown material. If the light is incident at 35 and refracts at 23, what is the in

Nastasia [14]

The index of refraction of the unknown material in which a ray of light is incident at 35° and refracted at 23° is 2.03

<h3>Snell's law</h3>

index of refraction (n) = Sine i / Sine r

n = Sine i / Sine r

Where

i is the angle of incidence
r is the angle of refraction

<h3>How to determine the refractive index </h3>

From the question given above, the following data were obtained:

Index of refraction of oil (nₒ) = 1.38
Angle of incidence (i) = 35°
Angle of refraction (r) = 23°
Index of refraction of unknown material (nᵣ) =?

nₒSine i = nᵣSine r

1.38 × Sine 35 = nᵣ × Sine 23

Divide both side by Sine 23

nᵣ = (1.38 × Sine 35) / Sine 23

nᵣ = 2.03

Thus, the index of refraction of the unknown material is 2.03

Learn more about Snell's law:

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7 0

2 years ago

Jonathan is pushing a 25 kg box with a force of 135 N. The friction opposing the motion is unknown. The overall net force for th

disa [49]

The magnitude of the frictional force of the box is 65 N

From the question given above, the following data were obtained:

Mass (m) = 25 Kg
Force applied (Fₐ) = 135 N
Net force (Fₙ) = 70 N
Frictional force (Fբ) =?

The frictional force acting on the box can be obtained as follow:

Fₙ = Fₐ – Fբ

70 = 135 – Fբ

Collect like terms

70 – 135 = – Fբ

–65 = –Fբ

Multiply through by –1

Thus, the frictional force of the box is 65 N

Learn more on Frictional force: brainly.com/question/25767576

8 0

2 years ago

point) A tank in the shape of an inverted right circular cone has height 5 meters and radius 4 meters. It is filled with 4 meter

zysi [14]

Answer:

W = 907963.50 J = 907.96 J

Explanation:

Note: Refer to the figure attached

Now, from the figure we have similar triangles ΔAOB and ΔCOD

we have

$\frac{5}{4}=\frac{x}{r}$

$r=\frac{4x}{5}$

Now, the work done to empty the tank can be given as:

$W = \int\limits^4_0 {(5-x)\rho\times g A} \, dx$

$W = \int\limits^4_0 {(5-x)1080\times 9.8 (\pi r^2)} \, dx$

$W = \int\limits^4_0 {(5-x)\times10584\times (\pi (\frac{4x}{5})^2)} \, dx$

$W = 6773.76\pi\int\limits^4_0 {(5-x)x^2)} \, dx$

$W = 6773.76\pi[\frac{5}{3}x^3-\frac{1}{4}x^4]^4_0$

$W = 6773.76\pi[\frac{128}{3}]$

W = 907963.50 J = 907.96 J

7 0

3 years ago

The thermodynamic efficiency of a heat engine that rejects heat at a rate of 20 mw when heat is supplied to it at a rate of 60 m

Alona [7]

The thermodynamic efficiency of a heat engine that rejects heat at a rate of 20 mw when heat is supplied to it at a rate of 60 mw is 66.67%.

It is given that,

Rejected heat at the rate of 20 mw and supplied heat a rate of 60 mw.

It is required to find the thermodynamic efficiency of a heat engine.

<h2>What is the thermodynamic efficiency of a heat engine that rejects heat at a rate of 20 mw when heat is supplied to it at a rate of 60 mw?</h2>

As we know that

Work Done = Heat Supplied - Heat Rejected

Work Done = 60 mw - 20 mw

Work Done = 40 mw

So the efficiency of heat engine will be:

$\eta=\frac{Work done}{Heat energy supply}$

$\eta=\frac{40}{60}$ × $100\\$

η = $\frac{2}{3}$ × $100\\$

η = 66.67%

Thus, thermodynamic efficiency of a heat engine that rejects heat at a rate of 20 mw when heat is supplied to it at a rate of 60 mw is 66.67%.

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#SPJ4

5 0

2 years ago