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likoan [24]
3 years ago
13

a student drops an object from the top of a building which is 19.6 m high. How long does it take the object to fall to the groun

d?
Physics
2 answers:
zubka84 [21]3 years ago
8 0

Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before. 
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.

This formula tells how far an object travels in how much time,
when it's accelerating:

               Distance = (1/2 acceleration) x (Time²).

                           D = 1/2 A T²

For your student who dropped an object out of the window,

     Distance = 19.6 m
     Acceleration = gravity = 9.8 m/s²

                                              D = 1/2 G T²

                                          19.6 =   4.9   T²

Divide each side by 4.9 :       4  =           T²

Square root each side:           2  =          T

When an object is dropped in Earth gravity,
it takes  2  seconds to fall the first 19.6 meters.

Crazy boy [7]3 years ago
5 0
Earth's gravity is 9.807 m/s²

Using this we can divide 19.6/9.807 to get 1.999 seconds or two if you round.
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2 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
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Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

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