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3 years ago

Which three statements about electromagnetic radiation are true?

Physics

2 answers:

grandymaker [24]3 years ago

3 0

I think that numbers one, three, and four are true

Svetllana [295]3 years ago

3 0

<u>Answers</u>

1)Light is a kind of electromagnetic radiation.

3)Electromagnetic radiation has no mass.

4)Electromagnetic radiation can travel through a vacuum.

<u>Explanations</u>

Electromagnetic waves are waves that can travel through vacuum. that is they do not require material medium for propagation.

Light is in the electromagnetic spectrum.

Electromagnetic waves can travel through a long distance. For instance, radio waves are electromagnetic radiations that travel long distances.

Electromagnetic radiations carry energy with it but no mass.

Some of the electromagnetic radiations are visible such as visible light.

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Acceleration = v^2 / r

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3 years ago

15. Use the chemical equation below to determine how many moles of ammonia

Anastaziya [24]

Answer:

I'm not really sure but I think it is

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3 years ago

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

3 0

3 years ago

Which ion channels mediate the falling phase of an action potontial?

White raven [17]

Answer:

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3 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

uysha [10]

Answer:

F=5449 N

Explanation:

Work done is a product of force and displacement ie

Work done, W, = Force*Displacement

Power, P, is Work done/Time

$P=\frac {W}{t}=\frac {FS}{t}$ where P is power, W is work done, F is force, S is displacement and t is time

In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328 W

Since displacement/time is velocity, then

P=FV where V is velocity in m/s

Making F the subject

$F=\frac {P}{V}$

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F=5449 N

7 0

3 years ago