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enyata [817]
3 years ago
5

If i apply 280 n of force to a 40kg object, what will it's acceleration be?

Physics
1 answer:
nika2105 [10]3 years ago
6 0

Answer:

f=ma

f=280N

m=40kg

a=?

280=40a

a=280/40

a=70N/kg

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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial
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3 years ago
Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for
Lelechka [254]

Answer:

The distance is  D  =  0.000712 \ m

Explanation:

From the question we are told that

    The wavelength of  the  light source is  \lambda  =  700 \ nm = 700 *10^{-9} \  m

     The distance from a pin hole is  x  =  9\ m

       The  diameter of the pin  hole is  d =  1.2 \ mm  =  0.0012 \ m

     

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              D  =  \frac{1.22 \lambda }{d }

substituting values

             D  =  \frac{1.22 * 700 *10^{-9} }{ 0.0012 }

             D  =  0.000712 \ m

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