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lesya [120]
2 years ago
9

A floating balloon can be formed when the substance helium is released from a compressed container into a flat rubber balloon. T

hen, the decompressed helium atoms expand and float up, making the rubber balloon expand around them and float with them. What is helium’s state when it does this? liquid solid gas plasma
Physics
2 answers:
mel-nik [20]2 years ago
6 0

Helium’s state when the decompressed helium atoms expand and float up, making the rubber balloon expand around them and float with them is gas.

 

<span>Helium is a chemical element with symbol He and atomic number 2. It is a colorless, odorless, tasteless, non-toxic, inert, monatomic gas, the first in the noble gas group in the periodic table. Its boiling point is the lowest among all the elements.</span>
Delicious77 [7]2 years ago
3 0
<h3><u>Answer;</u></h3>

Gas

<h3><u>Explanation;</u></h3>
  • The helium is in gaseous state.
  • Helium is an element that exists in gaseous state, it belongs to group 8 in the periodic table called noble gases.
  • A gas balloon is a balloon which flies since it is filled with a gas which is less dense than air or lighter than air. Such gases include; helium gas and hydrogen gas.
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Consider a simple ideal Rankine cycle with fixed turbine inlet conditions. What is the effect of lowering the condenser pressure
mr Goodwill [35]

Answer:

The effect of lowering the condenser pressure on different parameters is explained below.

Explanation:

The simple ideal Rankine cycle is shown in figure.

Effect of lowering the condenser pressure on

(a). Pump work input :- By lowering the condenser pressure the pump work increased.

(b) Turbine work output :- By lowering the condenser pressure the turbine work increased.

(c). Heat supplied :- Heat supplied increases.

(d). Heat rejected :- The heat rejected may increased  or decreased.

(e). Efficiency :- Cycle  efficiency is increased.

(f). Moisture content at turbine exit :- Moisture content increases.

8 0
1 year ago
10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

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

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

Learn more about work:

brainly.com/question/6763771

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4 0
2 years ago
Before you start taking measurements though, we’ll first make sure you understand the underlying concepts involved. By what meth
Svetradugi [14.3K]

Answer:

If they are metallic spheres  they are connected to earth and a charged body approaches

non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.

If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.

8 0
2 years ago
An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v
Aneli [31]

We are given: Final velocity (v_f)=20 m/s .

Time t= 2.51 s and

distance s = 82.9 m.

We know, equation of motion

v_f = v_i + at.

Let us plug values of final velocity, and time in above equation.

20=v_i+a(2.51)

20=v_i+2.51a

Subtracting 2.51a from both sides, we get

20-2.51a=v_i  -----------equation(1)

Using another equation of motion

v_f-v_i=2as

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

20-(20-2.51a)=2*a(82.90)

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.


5 0
1 year ago
How do you do projectile motion problems
PtichkaEL [24]
The key  projectile motion is that gravity allows downward only
7 0
2 years ago
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