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3 years ago

7

Properties of helium

1 answer:

postnew [5]3 years ago

7 0

Answer:

Helium has many unique properties: low boiling point, low density, low solubility, high thermal conductivity and inertness, so it is use for any application which can explioit these properties. Helium was the first gas used for filling balloons and dirigibles

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A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a

Burka [1]

Answer:

350 N/m

Explanation:

If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m

k = 70 / 0.2 = 350 N/m

The spring constant is 350 N/m

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3 years ago

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*Materials that regulate the flow of current through them *

4vir4ik [10]

Answer:

electromagnet

Explanation:

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A ball starts from rest. It rolls down a ramp and reaches the ground after 4 seconds. Its final velocity when it reaches the gro

Hunter-Best [27]

If it starts at rest the initial velocity is 0.
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3 years ago

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LUCKY_DIMON [66]

Answer:

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3 years ago

wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

$E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)$

The electric field due to charge QB at P is EB leftwards

$E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)$

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0

3 years ago