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gtnhenbr [62]
3 years ago
12

A sample of helium behaves as an ideal gas as energy is

Physics
1 answer:
alisha [4.7K]3 years ago
4 0

Answer:

0.0321 g

Explanation:

Let helium specific heat c_h = 5.193 J/g K

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

E_h = m_hc_h \Delta T = 20 J

where m_h is the mass of helium, which we are looking for:

m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g

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Why is water considered to be a complex compound
DedPeter [7]
HEY THERE !!

\huge{COMPOUND:-}
Compounds contain two or more different elements. 


Water is a molecule because it contains molecular bonds . Water is also a compound  because it is made from more than one kind of element. (oxygen and hydrogen).

So If you like, you can say that water is a molecular compound.

HOPE IT HELPED YOU.
7 0
3 years ago
Read 2 more answers
Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a len
Alenkinab [10]

Answer:

6.75J

Explanation:

U=1/2KΔx²

U=0.5* 150*0.30^2

4 0
2 years ago
What cloud extending to the gravitational limits of the solar system would comets come from?.
ser-zykov [4K]

The Oort cloud extends to the gravitational limits of the solar system would comets come from.

<h3>What is oort cloud?</h3>

It is a hypothetical idea of a cloud of mostly frozen planetesimals that would orbit the Sun at distances between 2,000 and 200,000 AU.

The Oort cloud reaches the solar system's outer gravitational boundaries, where comets originate.

Hence cloud extending to the gravitational limits of the solar system will be oorto cloud.

To learn more about the oort cloud refer;

brainly.com/question/23368033

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4 0
2 years ago
Who can help me?? physic question​
IgorC [24]

Assuming acceleration due to gravity of the moon is constant and there’s no initial velocity in the mans jump we can use one of the kinematic equations. x(final)=x(initial)+(1/2)gt^2. Plug in known values. 0=10-(1.62/2)t^2. The value 1.62 is acceleration of gravity on the moon. Now simply solve for t. t=3.513

5 0
3 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
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