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3 years ago

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Larry sees a group of people weeping, with frowns on their faces and their eyes turned down. Larry their expressions to understa

nd that they are feeling sadness.
Select one:
a. misinterprets
b. decodes
C. encodes X
d. recodes

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

c

Explanation:

no need explanation u can trust me

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1)the magnetic field is strongest near_________of a bar magnet.

horrorfan [7]

1. Is A. at the poles because thats where the magnetic field is going out then coming back into the earth to produce the magnetic field.
2. Again its A. because the compass needle is attracted to " north " which is magnetic south. It does this because opposites attract.
3. This one would be B. Because if the magnets were being repelled the magnetic field lines would look like there was a line that the field hit and bounced off of it.
4. This answer is A. the magnetite helps them migrate so they know which way is north and which way is south.
5. This answer is A. Because without the domains there wouldn't be poles on the magnetic object. <span />

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3 years ago

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3) A tolley of mass 4kg, moving with a velocity

Leokris [45]

Answer:

3.57 m/s

Explanation:

The sum of the 2 momentums Is equal the finale momentums. so if momentums Is q, v Is velocity and m Is Mass, q3=m1*v1+m2**v2=16+9=25 m*kg/s

q3=m3*v3

v3=q3/m3=25/(4+3)=3.57m/s

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3 years ago

Why does your hand feel cool if you spill some alcohol on it?

son4ous [18]

<span>It takes heat to make something evaporate, so it takes heat from your arm. Alcohol easily evaporates at room temperature, so it feels cool. This is also why you feel cool when getting out of the pool. The water on your skin evaporates.

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3 years ago

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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

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4 0

3 years ago

Two particles collide, one of which was initially moving and the other initially at rest. Is it possible for both particles to b

solmaris [256]

Answer:

Not possible

Explanation:

Unless there's some extra external force to keep both particles at rest after the collision, the momentum must be conserved before and after the collision.

So before the collision, 1 particle is at rest, 1 not -> total momentum is non-zero

After the collision, both particles are at rest -> total momentum is zero which is different from before.

Therefore this is not possible.

4 0

3 years ago