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2 years ago

14

At this rate, what fraction of its mass would io lose in 4. 5 billion years?.

1 answer:

Pie2 years ago

4 0

In 4.5 billion years, the amount of mass lost is 1.41912 * 10¹⁷ kg.

<h3>Equation</h3>

An equation is an expression used to show the relationship between two or more variables and numbers.

Given that:

Io loses about a ton (1000 kilograms) of sulfur dioxide per second to Jupiter's magnetosphere.

In 4.5 billion years, amount of mass lost = 1000 kg/s * (4.5 * 10⁹ * 365 * 60 60 * 24) = 1.41912 * 10¹⁷ kg.

In 4.5 billion years, the amount of mass lost is 1.41912 * 10¹⁷ kg.

Find out more on Equation at: brainly.com/question/13763238

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What is the english system of measurement called?

Artemon [7]

The answer is imperial units

8 0

3 years ago

If the normal force exerted by the track on the car when it is at the top of the track (point B) is 6.00 N , what is the normal

Eddi Din [679]

Answer:

So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.

Now,

We have F=ma,

and a=v/t,

so we can have another equation for it as,

Now, providing the required data to, it; ∴t =2 sec,

F=(1)×(20/2),

<u>F=10 N.</u>

So, the car would be acting the force,F of about 10 N while the car is present on the lower region of the track.

4 0

3 years ago

Imagine you use Nitrogen as your gas. If you have the cold side as cold as you can without liquefying it (78 K), and run the hot

alina1380 [7]

Answer:

The efficiency of a Stirling engine is 74%

Explanation:

Given:

Temperature of gas when it is cold $T_{1} = 78$ K

Temperature of gas when it is hot $T_{2} = 300$ K

The efficiency of a stirling engine,

$\eta =1 - \frac{T_{1} }{T_{2} }$

$\eta = 1- \frac{78}{300}$

$\eta = 1-0.26$

$\eta = 0.74$

∴ $\eta = 74 \%$

Therefore, the efficiency of a Stirling engine is 74%

5 0

3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of 0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = <em>0.27</em>

7 0

3 years ago

Trying to beat the heat of the last summer, a physics grad student went to the local toy store and purchased a plastic child's s

svp [43]

Answer:

625 piece.

Explanation:

Let n be the required no of piece

200 litre = 200 x 10⁻³ m³ = 200 x 10⁻³ x 10³ kg = 200 Kg

mass of ice piece = n x 30 x /1000

Heat lost by ice pieces = (n x 30) / 1000 x ( 80 + 1 x 16⁰C)

Heat gained by water = 200 x 1 x 9⁰C

Heat lost = Heat gained

(n x 30) /1000 x ( 80 + 16 ) =200 x 9

2.88 n = 1800

n = 625

6 0

4 years ago