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schepotkina [342]
3 years ago
12

Determine the number of hydrogen atoms in the Sun. The mass of the Sun is 2×1030kg. About 70% of it by mass is hydrogen, 30% is

helium, and there is a negligible amount of other elements.
Physics
1 answer:
valentinak56 [21]3 years ago
8 0

Answer:

8.36541\times 10^{56}

Explanation:

Mass of the sum = 2\times 10^{30}\ kg

70% of this mass is

0.7\times 2\times 10^{30}=1.4\times 10^{30}\ kg

Mass of hydrogen in the sun is 1.4\times 10^{30}\ kg

Mass of a hydrogen atom = 1.6735575\times 10^{-27}\ kg

Dividing the mass hydrogen in the sun by the mass of one hydrogen atom will give us the number of hydrogen atoms

\frac{1.4\times 10^{30}}{1.6735575\times 10^{-27}}=8.36541\times 10^{56}

The number hydrogen atoms in the Sun is 8.36541\times 10^{56}

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Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe a
masha68 [24]
Missing question: "What is the spring's constant?"

Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m
And by using Hook's law, we can find the constant of the spring:
k= \frac{F}{\Delta x}= \frac{67.6 N}{0.096 m}=704.2 N/m
4 0
3 years ago
The most reactive metals are located at the of the periodic table.
NeX [460]

Answer:

the elements towards the bottom left corner

7 0
3 years ago
Read 2 more answers
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
Please help.. I'm desperate.
rosijanka [135]

Answer:

Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry.

Explanation:

7 0
2 years ago
A 3.5-kg object placed on an inclined plane (angle 30? above the horizontal) is connected by a string going over a pulley to a 1
Blababa [14]

Answer:

a= 0.22 m/s²

Explanation:

Given that

M = 3.5 kg

θ = 30°

m = 1 kg

μ= 0.3

The force due to gravity

F₁= M g sinθ

F₁=3.5 x 10 x sin 30

F₁= 17.5 N

F₂ = m g

F₂ = 1 x 10 = 10 N

The maximum value of the friction force on the incline plane

Fr = μ M g cosθ

Fr = 0.3 x 2.5 x 10 cos30°

Fr= 6.49 N

Lets take acceleration of the system is a  m/s²

F₁ - F₂  - Fr =  (M+m) a

17.5 - 10 - 6.49 = (3.5+1)a

a= 0.22 m/s²

7 0
2 years ago
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