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3 years ago

A fragment of a collapsing gas cloud that comes to equilibrium with a central temperature of 4 million k will become a:

Physics

1 answer:

Liono4ka [1.6K]3 years ago

4 0

If there's enough Hydrogen in the cloud, then fusion reactions
will ignite in the core, and it'll become a star.

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Why mole is called fundamental unit.

gladu [14]

Explanation:

because it doesn't depend upon other unit like kg meter and second

4 0

3 years ago

Rama's weight is 40kg. She is carrying a load of 20 kg up to a height of 20 m . What work does she do?

Sliva [168]

Answer:

$\huge\star{\underline{\mathtt{\blue{Answer}}}}\huge\star$ ...

$\huge\boxed{\fcolorbox{white}{blue}{mgh=4000}}$

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7 0

3 years ago

A 1.50 kg rock is thrown up into the air from ground level, reaches a maximum height of 7.00 m, then returns to the ground. Calc

Minchanka [31]

Answer:

17.565 kgm/s

Explanation:

Momentum = mass × velocity

I = mv..................... Equation 1

But we can calculate the value of v using the equation of motion under gravity.

v² = u²+2gs............. Equation 2

Where u = initial velocity, s = maximum heigth, g = acceleration due to gravity.

Given: u = 0 m/s (at the maximum heigth), s = 7.0 m.

Constant: g = 9.8 m/s²

Substitute these values into equation 2

v² = 0²+ 2×7×9.8

v² = 137.2

v = √137.2

v = 11.71 m/s.

Also given: m = 1.50 kg

substitute these values into equation 1

Therefore,

I = 1.5×11.71

I = 17.565 kgm/s

3 0

3 years ago

A car has a mass of 1,200 kg. What is its acceleration when the engine exerts a force of 600 N? (Formula: F=ma)

Rasek [7]

Answer:

Option (a)

Explanation:

Given that,

Mass of a car, m = 1200 kg

Force exerted by the engine, F = 600 N

Noe force,F = ma

a is the acceleration of the engine

$a=\dfrac{F}{m}\\\\a=\dfrac{600\ N}{1200\ kg}\\\\a=0.5\ m/s^2$

So, the acceleration of the car is 0.5 m/s².

6 0

3 years ago

An isloated point charge produce an electric field with magnitude E at a point 2 m away. At a point 1 m from the charge magnitud

Nina [5.8K]

Answer:

the correct answer is C, E’= 4E

Explanation:

In this exercise you are asked to calculate the electric field at a given point

E = $k \frac{q}{r^2}$

indicates that the field is E for r = 2m

E = $\frac{ k q}{4}$ (1)

the field is requested for a distance r = 1 m

E ’= k \frac{q}{r'^2}

E ’= k q / 1

from equation 1

4E = k q

we substitute

E’= 4E

so the correct answer is C

8 0

3 years ago