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Katyanochek1 [597]
3 years ago
9

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr

iction is absent. The spring constant of the spring is 4900 N/m. The piston has a negligible mass and a radius of 0.029 m. (a) When the air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress
Physics
1 answer:
Mazyrski [523]3 years ago
6 0

Answer:

x = 0.0537 m or 5.37 cm

Explanation:

Given:

spring constant'k'= 4900 N/m

radius 'r' =0.029 m

Area 'A' =r²π = 0.029²π => 2.6 x 10^{-3} m²

Here, Pressure 'P' is given by,

Pressure = Force / Area

And we know that, for a spring :

F = kx, where k is the spring constant and x is the change in length.

P = kx/A

As P = 101325 Pa

101325 = 4900x / ( 2.6 x 10^{-3})

x = 0.0537 m or 5.37 cm

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A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while
zlopas [31]

Answer:

new energy of the capacitor is half that of initial energy

Explanation:

As we know that the energy stored in the capacitor is given as

U = \frac{1}{2}CV^2

here we know that capacitance of parallel plate capacitor is given as

C = \frac{\epsilon_0 A}{d}

V = voltage of battery

so now we have capacitor remains connected to the same battery and the separation between the plates is doubled

so we have

C' = \frac{\epsilon_0 A}{2d}

so now the energy stored between the plates is

U = \frac{1}{2}(\frac{C}{2})V^2

so new energy of the capacitor is half

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4 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

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7 0
3 years ago
Three crates with various contents are pulled by a force F pull = 3615 N across a horizontal, frictionless roller‑conveyor syste
Dmitry_Shevchenko [17]

Answer:

m1=914.9kg

m2=604.9kg

m3=864.75kg

Explanation

I think we are suppose to find the mass of the crate.

The effective force that moves the body in positive x direction is 3615N

ΣFx = Σma

Then Fx=3615N

Then the masses be m1, m2 and m3

Then,

ΣF = Σ(ma)

3615=(m1+m2+m3)a

Given that a=1.516

The masses are

m1+m2+m3=, 2384.56. Equation 1

Between mass 1 and mass 2 is, F12=1387.

The effective force that pull mass 1 is 1387.

F12=m1 ×a

Therefore,

m1=F12/a

m1=1387/1.516

m1=914.9kg.

The effective force that pulls crate 1 and crate 2 is F23

F23=(m1+m2)a

Therefore

2304=(m1+m2)a

Therefore, since a=1.516

m1+m2=2304/1.516

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Since m1=914.9kg

So, m2=1519.8-m1

m2=1519.8-914.9

m2=604.9kg

Also from equation 1

m1+m2+m3=2384.56

Since m1=914.9kg and m2=604.9kg

Then, m3=2384.56-604.9-914.9

m3=864.75kg

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