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Rashid [163]
3 years ago
7

A thin-walled vessel of volume V contains N particles which slowly leak out of a small hole of area A. No particles enter the vo

lume through the hole. Find the time required for the number of particles to decrease to N/2. Express your answer in terms of A, V, and v.
Physics
1 answer:
LuckyWell [14K]3 years ago
7 0

Answer:

    \frac{V}{2av}

Explanation:

From the question we are told that

Volume V

Contains N particles

Leaks from a small hole of area A

Generally the equation for Flow rate is given as

Volume Flow Rate  V_r = A * v

Mathematically we find the  time taken to flow half way which is given by

          \frac{(V/2)}{A*v}

Therefore the  time taken is

           \frac{V}{2av}

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a wrench weighs 5.24 newtons on earth. when it is taken to the Moon, where g =1.16 m/s2 how much does it weigh?
Andrew [12]

“Weight of the wrench” on “the moon” is “6.07 kg”.

<u>Explanation</u>:

Weight of the wrench is 5.24 N  

Weight of the wrench in kilograms = W × g

Taken “g” on the moon is 1.16 \mathrm{m} / \mathrm{s}^{2}

=5.24 \mathrm{N} \times 9.81 \mathrm{m} / \mathrm{s}^{2}=51.352 \mathrm{kg}

Weight of the wrench in kilograms is 51.352 kg.

Formula to calculate weight of the object on the moon is

\frac{\text {weight of the object on earth}}{9.81 \mathrm{m} / \mathrm{s}^{2}} \times 1.16 \mathrm{m} / \mathrm{s}^{2}

Substitute the values given,

=\frac{51.352 \mathrm{kg}}{9.81 \mathrm{m} / \mathrm{s}^{2}} \times 1.16 \mathrm{m} / \mathrm{s}^{2}

=5.234 \times 1.16

= 6.07 kg

Therefore, weight of the wrench on the moon is 6.07 kg.  

6 0
3 years ago
Lamar writes several equations trying to better understand potential energy. What conclusion is best supported by lamar's worm?
kumpel [21]

Complete question is;

Lamar writes several equations trying to better understand potential energy. What conclusion is best supported by Lamar’s work?

A) The elastic potential energy is the same for any distance from a reference point.

B) The gravitational potential energy equals the work needed to lift the object.

C) The gravitational potential energy is the same for any distance from a reference point.

D) The elastic potential energy equals the work needed to stretch the object

Answer:

B) The gravitational potential energy equals the work needed to lift the object.

Explanation:

In physics, we know that potential energy is the energy of a body at rest while the energy of a body in motion is known as kinetic energy.

However,the work required to lift a body from it's position of rest is equal to the Gravitational potential energy of that body.

Elastic potential energy is the one that is stored as a result of force applied to deform an elastic object. Thus, it is not equal to the work needed to stretch the object and it is also not the same for any distance from reference point.

Thus, looking at the options, Option B is correct

6 0
3 years ago
Observing neutrinos from the Sun is an important way to check the fusion rate, but it can be very tough to build a machine that
Vika [28.1K]

Answer:

The trouble that the most recent experiment, Borexino, have to overcome was that

neutrinos hardly interact with matter and so radioactive decay of ant material inside the detector could look exactly like a neutrino interaction too

5 0
3 years ago
What is the velocity of the buggy? <br><br> At 20 seconds the buggy will have a position of ____ ?
lbvjy [14]
At 20 seconds it will be 12.6 because at 10 seconds it was as at approximately 6.3 so we times it by 2 to get the 20s
6 0
3 years ago
Contrast the force of gravity between these pairs of objects a 1 kg mass and a 2 kg mass that are 1 m apart and two 2 kg masses
Yuki888 [10]

Solution

Force between pair of objects of masses 1kg and 2kg that are 1m apart is given as

F=G\times \frac{1\times 2}{1^2}

here G is gravitational constant

G=6.674\times10^{-11}Nm^2kg^{-2}

therefore,

F=6.674\times10^{-11}Nm^2kg^{-2}\times 2

F=13.348\times10^{-11}Nm^2kg^{-2}

similarly Force between pair of objects of masses 2kg each that are 1m apart is given as

F'=G\times \frac{2\times 2}{1^2}

F'=6.674\times10^{-11}Nm^2kg^{-2}\times 4

F'=2\times 13.348\times10^{-11}Nm^2kg^{-2}

or F'=2F

it means  Force between pair of objects of masses 2kg each that are 1m apart is equal to twice the Force between pair of objects of masses 1kg and 2kg that are 1m

7 0
3 years ago
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