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vovangra [49]
3 years ago
12

Assuming that each of the following objects is a typical example of its class, rank them by increasing density.

Physics
1 answer:
inysia [295]3 years ago
8 0

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>.

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A 75-kilogram hockey player is skating across the ice at a speed of 6.0 meters per second. What is the magnitude of the average
saveliy_v [14]

Answer:

(3) 690 N

Explanation:

Force: This is the product of mass and acceleration of a body. The S.I unit of force is Newton(N).

The formula for force is given as,

F = ma..................... Equation 1

Where F = force, m = mass, a = acceleration.

Also,

a = (v-u)/t................... Equation 2

Where v = Final velocity, u = initial velocity, t = time.

Given: u = 6.0 m/s, v = 0 m/s (bring to stop), t = 0.65 s.

Substitute into equation 2

a = (0-6)/0.65

a = -6/0.65

a = -9.23 m/s²

Also given: m = 75 kg

Substitute again into equation 1

F = 75(-9.23)

F = -692.25 N

The negative sign tells that the force oppose the motion of the player

F ≈ 690 N

Hence the right option is (3) 690 N

3 0
3 years ago
A displacement vector has a magnitude of 810 m and points at an angle of 18° above the positive x axis. what are the x and y sca
Naily [24]
I attached my work and highlighted my answers. Hope it makes sense! please comment back with any questions if anything is still unclear! :)

5 0
3 years ago
The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a
vampirchik [111]

Answer:

F = 1.5 \times 10^{-16} N

this force is 1.68 \times 10^{13} times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

KE = 1 keV

KE = 1 \times 10^3 (1.6 \times 10^{-19}) J

KE = 1.6 \times 10^{-16} J

now the speed of electron is given as

KE = \frac{1}{2}mv^2

now we have

v = \sqrt{\frac{2 KE}{m}}

v = 1.87 \times 10^7 m/s

now the maximum force due to magnetic field is given as

F = qvB

F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})

F = 1.5 \times 10^{-16} N

Now if this force is compared by the gravitational force on the electron then it is

\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}

\frac{F}{F_g} = 1.68 \times 10^{13}

so this force is 1.68 \times 10^{13} times more than the gravitational force

4 0
3 years ago
If a car is moving to the left with constantvelocity, one can conclude thatthere mustbe no forces applied to the car.the netforc
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Answer:

the net force applied to the car is zero.

Explanation:

According to Newton's second law, the acceleration of an object (a) is directly proportional to the net force applied (F):

a=\frac{F}{m}

where m is the object's mass.

In this problem, the car is moving with constant velocity: this means that the acceleration is zero, a = 0. Therefore, according to the previous equation, the net force must also be zero: F = 0. So, the correct answer is

the net force applied to the car is zero.

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