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ludmilkaskok [199]
2 years ago
14

PLEASE HELP QUICKLY!!!!!

Physics
1 answer:
Grace [21]2 years ago
3 0

Answer:

D air

Explanation:

it is not found on the periodic table

brainliest plsssssssssssssssss

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A banana peel has lots of friction.<br> True or False
olga_2 [115]

Answer:

False

Explanation:

I learned it the hard way trust me T^T

3 0
2 years ago
Define the term work​
bagirrra123 [75]

Answer:

Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force

Explanation:

at least part of which is applied in the direction of the displacement. ... To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd.

5 0
2 years ago
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Assuming that each of the following objects is a typical example of its class, rank them by increasing density.
inysia [295]

molecular cloud <interstellar cloud <1 Msun protostar <1 Msun star <intercloud gas

Explanation:

<u>Molecular cloud-</u> They are a variety of interstellar cloud in which molecular hydrogen can sustain themselves. They have a very low temperature ranging from -440 to -370 degrees Fahrenheit or between<u> 10 to 50 Kelvin. </u>Owing to their extremely low temperature, they appear mostly dark when viewed through telescopes.

<u>Interstellar cloud-</u> They are a congregation of a large number of interstellar gases, dust and plasma in any galaxy or universe. They have varying temperature depending on their proximity to a star. E.g. Neutral hydrogen atom clouds have a temperature of around <u>just 100 Kelvin</u> while those in the near vicinity of a star have temperatures as high as 10,000 Kelvin.

<u>1 Msun star-</u> These stars have temperature anywhere between <u>5300 and 6000 Kelvin</u>. The main source of such high surface temperature is nuclear fusion process where elemental hydrogen molecules are fused to form helium molecules.  

<u>1 Msun protostar-</u> protostar is rather a young star which is still in formation phase (i.e. gathering mass from the parent molecular cloud). They have temperature anywhere between <u>2000-3000</u> kelvin and are accompanied by dust usually.

<u>Intercloud gas- </u>These are the remainder gases that are spread throughout the interstellar space. This Intercloud gas is divided into warm intercloud medium and extremely hot coronal gas with temperatures comparing to Sun’s corona. Warm intercloud forms the dominant part of intercloud gas with a temperature around <u>8000 Kelvin</u>.

8 0
3 years ago
Convert the following measurements. Only type numbers and decimals into the answer boxes:
ivolga24 [154]

Answer:

2m 4m2Explanation:rteji3ujnrej

6 0
2 years ago
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Consider two identical objects released from rest high above the surface of Earth. (Neglect air resistance for this question.)
Nadusha1986 [10]

The question is missing its alternatives. Here is the complete question.

Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.

1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.

a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;

Answer: C) K2 = (4/3)K1

Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - \frac{G.Me.m}{r} + U₀

In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.

The potencial energy of an object of mass m on the surface of the Earth is:

Usurface = - \frac{G.Me.m}{Re}

The potencial energy of the object in Case 1 is

U1 = - \frac{G.Me.m}{2Re}

For the Case 2:

U2 = - \frac{G.Me.m}{3Re}

The potencial change in Case 1:

ΔU1 = - G.Me.m.(\frac{1}{Re}-\frac{1}{2Re}) = - \frac{1}{2}\frac{G.Me.m}{Re}

For Case 2:

ΔU2 = - G.Me.m(\frac{1}{Re}-\frac{1}{3Re}) = - \frac{2}{3}\frac{G.Me.m}{Re}

Comparing ΔK1 and ΔK2 equals comparing ΔU1 and ΔU2:

Δ\frac{U2}{U1} = (-2/3)(-1/2) = 4/3

So, comparing kinetic energies, K2 is 4/3 of K1.

5 0
3 years ago
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