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2 years ago

4. Interference is an example of which aspect of electromagnetic radiation?

Physics

2 answers:

Lisa [10]2 years ago

6 0

Answer:

D is the answer wave behavior

soldi70 [24.7K]2 years ago

4 0

Interference, refraction, diffraction, and dispersion are all aspects of wave behavior. (D). That is, particles don't do these things.

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Normally the rate at which you expend energy

Greeley [361]

Answer:

80 min

Explanation:

Distance = rate × time

280 Cal = 3.5 Cal/min × t

t = 80 min

4 0

2 years ago

At 45 degrees s latitude, the angle of the noon sun is lowest and the length of daylight is shortest on:

Marat540 [252]

Answer;

- June 21

At 45 degrees latitude, the angle of the noon sun is lowest and the length of daylight is shortest on June 21.

Explanation;

-On June 21 you will note that the Northern Hemisphere is pitched toward the Sun. This means that the Sun's vertical ray is striking the Earth at the Tropic of Cancer (23.5 degrees N).

-Days tend to get longer in the northern hemisphere from December 21 to June 21, and then grow shorter from June 21 to December 21. The June solstice is the summer solstice in the Northern Hemisphere and the winter solstice in the Southern Hemisphere. The date varies between June 20 and June 22, depending on the year, and the local time zone.

8 0

3 years ago

What is another word that can be used to describe the position of the object?

steposvetlana [31]

where the object is located

6 0

2 years ago

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizo

Andreyy89

Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?

Answer:

$W_{work}=2.67*10^{-3}J$

Explanation:

Given data

Charge q=28 nC

Electric field E=5.00×10⁴ V/m.

Distance d=2.70 m

Angle α=45°

To find

Work done by electric force

Solution

$W_{work}=F_{force}*D_{distance}Cos\alpha \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J$

8 0

3 years ago

Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l

OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

$Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\$

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

$F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\$

F = 85696.5 N = 85.69 KN

8 0

2 years ago