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3 years ago

if a spring has a spring constant of 2 N/m and it is stretched 5 cm, what is the force of the spring?

Physics

1 answer:

djyliett [7]3 years ago

4 0

Answer:

0.1 N

Explanation:

Considering the relationship between force,

spring constant and extension as defined by Hook's law

The force F=xk as from Hooke's law where F is the force of the spring, k is spring constant and x is extension or compression. Substituting 2 N/m for k and 5cm which is equivalent to 0.05 m for extention x then the force will be

F=2*0.05=0.1 N

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Answer:

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2 years ago

When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t

antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

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k = -0.4621

I = I(0)e^(-0.4621t)

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Intensity of beam 18 feet below the surface is about 0.2%

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3 years ago

A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow

shusha [124]

Answer:

150000000

$\dfrac{F_e}{F_g}=0.00000203873598369$

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

$1\ C=6.25\times 10^{18}\ electrons$

Number electrons is

$n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000$

The number of electrons added or removed was 150000000

Force is given by

$F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369$

The ratio is $\dfrac{F_e}{F_g}=0.00000203873598369$

Balancing the forces we get

$Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C$

The electric field required is 49050000 N/C

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4 years ago

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I think it is the wave energy.
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