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Ganezh [65]
3 years ago
14

If the pressure of a gas is really due to the random collisions of molecules with the walls of the container, why do pressure ga

uges – even very sensitive ones – give perfectly steady readings? Shouldn't the gauge be continually jiggling and fluctuating? Explain.
Physics
1 answer:
dusya [7]3 years ago
6 0

Answer:

there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

Explanation:

The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²

To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.

In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface

           # _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1

           #_molecular = 6.02 10¹⁷ molecules per second

therefore the variation of the number of molecules is not very important

Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

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A baseball is hit from a height of 4 feet above the ground with an initial velocity of 110 feet per second and at an angle of 35
mrs_skeptik [129]

Answer:

Part a)

h = 62.2 ft

Part b)

It will not able to cross the pole

Explanation:

As we know that ball is hit with speed of 110 ft/s at an angle of 35 degree

so here we will say

v_y = 110 sin35

v_x = 110 cos35

now at the maximum height the vertical velocity will become zero

so here we can use kinematics

v_f^2 - v_y^2 = 2 a h

here we have

a = -32 ft/s^2

v_f = 0

v_y = 63.1 ft/s

now we have

0 - 63.1^2 = 2(-32)h

h = 62.2 ft

Part b)

now the height of ball is related to the distance from point of projection is given as

y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}

now we know that

x = 375 ft

y = 375(tan35) - \frac{(32)(375^2)}{2(110^2)(cos35)^2}

y = 262.6 - 277.12 = -14.5 ft

since its coming negative so it will not able to cross the pole

3 0
3 years ago
An object moves in uniform circular motion at 25 m/s and takes 1.0 second to go a quarter circle. Calculate the centripetal acce
tekilochka [14]

Given:

Object in circular motion 25 m/s

1 second to go quarter circle

Required:

Centripetal acceleration:

Solution:

Acceleration = v2/r

Where v is the velocity and r is the radian

Substituting the values into the equation,

Acceleration = v2/r = (25 m/s)2/(4*pi/180) = 8952.47 m2/s2

4 0
4 years ago
If we compare the force of gravity too strong nuclear force we would conclude that
Yuri [45]
The force of gravity is much weaker than the strong nuclear force. But the strong nuclear force only acts over short distances, such as within the nuclues. The gravitational force can act over infinite distance.

6 0
4 years ago
Read 2 more answers
250 W 12 km/h<br>4. Convert yoor answer from 3 to meters per second.​
taurus [48]

Answer: 3.33 m/s

Explanation:

Assuming the questions is to convert  12 km/h to meter per second (m/s), let's begin:

In order to make the conversion, we have to know the following:

1 km=1000 m

And:

1 h=3600 s

Keeping this in mind, we can make the conversion:

12 \frac{km}{h} \frac{1000 m}{1 km} \frac{1h}{3600 s}

Then:

12 \frac{km}{h}= 3.33 \frac{m}{s}

5 0
4 years ago
Which is a characteritic of a physical change?
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Some thing that is done physically like with a persons hands and such
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