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Anastaziya [24]
3 years ago
5

Which of the following could be vector magnitudes?

Physics
2 answers:
borishaifa [10]3 years ago
8 0

Answer:

A and D

Explanation:

A vector magnitude has a magnitude and a direction Then 15 km are a magnitude that could have a direction for example 15 km in north direction.

The same happends with 30 m/s its a magnitude that could have a direction.

In the case of the option B 25 mililiters is a scalar magnitude because is a volumen then we can not define a direction, and the same for the option C 15 hours.

SOVA2 [1]3 years ago
5 0

Answer:

Both A and D

Explanation:

Vector magnintude contains both speed and direction and so do these answer choices of 15km and 30m/s

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The largest stars are _____ times the mass of the Sun. 10 100 1,000 10,000
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100 times

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You place your pencil on your desk it stays there
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Two students, Student X and Student Y, stand on a long skateboard that is at rest on a flat, horizontal surface, as shown. In or
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the answer is B.

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Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T
Zina [86]

a) Total power output: 3.845\cdot 10^{26} W

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is 540 W/m^2

Explanation:

a)

The intensity of electromagnetic radiation is given by

I=\frac{P}{A}

where

P is the power output

A is the surface area considered

In this problem, we have

I=1360 W/m^2 is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

r=1.5\cdot 10^{11} m (distance Earth-Sun)

Therefore

A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2

And now, using the first equation, we can find the total power output of the Sun:

P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

E=hf

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

P=\frac{E}{t}

where t is the time.

This means that the power output is proportional to the frequency:

P\propto f

Here the frequency increases by 1 MHz: the original frequency was

f_0 = 60 MHz

so the relative percentage change in frequency is

\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%

And therefore, the power also increases by 1.67 %.

c)

In this second  case, we have to calculate the new power output of the Sun:

P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m

Now we can find the radiation intensity with the equation

I=\frac{P}{A}

Where the area is

A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2

And substituting,

I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2

Learn more about electromagnetic radiation:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

4 0
3 years ago
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