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aliina [53]
3 years ago
8

It moved from 0 cm to 5 cm at a constant speed of 1 cm/s.

Physics
1 answer:
Stels [109]3 years ago
8 0

It moved from 0 cm to 4 cm at a constant speed of 1 cm/s.

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Two metal balls have charges of 7.1 × 10-6 coulombs and 6.9 × 10-6 coulombs. They are 5.7 × 10-1 meters apart. What is the force
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6 0
3 years ago
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A 24 liter tank contains ideal helium gas at 27 degrees C anda pressure of 22 atm. How many moles of gas are in the tank?
frozen [14]

Answer:

d)21.5 moles

Explanation:

Given that

L = 24 L

T = 27 °C = 300 K

P = 22 atm

We know that ideal gas equation

P V = n R T

P=Pressure ,V= Volume ,n=Moles ,R=Universal gas constant ,T=Temperature

Now by putting the values

22 x 24 = n x 0.08206 x 300

n= 21.447 moles

n= 21.5 moles

Therefore the number of moles will be 21.5 moles.

The answer is "d".

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3 years ago
What is hooke's law? does it apply to elastic materials or to inelastic materials?
zysi [14]
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6 0
4 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a cha
soldier1979 [14.2K]
The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854  *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854  *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854  *10 ^-12 equal to -1.908 *10^7. 

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8 0
3 years ago
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Prove(show) ''T=2π√(l/g)''​
Nonamiya [84]

Answer:

Time period for Simple pendulum, T=2\pi\sqrt{\frac{l}{g}

Explanation:

The Simple Pendulum

Consider a small bob of mass m is tied to extensible string of length l that is fixed to rigid support. The bob is oscillating in the plane about verticle.

       Let \theta is the angle made by string with vertical  during oscillation.

Vertical component of the force on bob, F=-mg\sin\theta

Negative sign shows that its opposing the motion of bob.

Taking \theta as very small angle then, \sin\theta\sim\theta

F=-mg\theta    

Let x is the displacement made by bob from its mean position ,

then, \theta=\frac{x}{l}

so, F=-mg\frac{x}{l}                ........(1)

Since, pendulum is in hormonic motion,

as we know, F=-kx

where k is the constant and k=m\omega^{2}

F=-m\omega^2x                   .........(2)

From equation (1) and (2)

-m\omega^2x=-mg\frac{x}{l}

\omega=\sqrt{\frac{g}{l}}

Since, \omega=\frac{2\pi}{T}

\frac{2\pi}{T}=\sqrt{\frac{g}{l}

T=2\pi\sqrt{\frac{l}{g}}

6 0
3 years ago
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