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Debora [2.8K]
4 years ago
8

A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g

round is 120.00 meters. Find the time the bottle would take to land if it were to fall the same distance on the moon instead of Earth. (Note: Acceleration due to gravity on the moon is 1/6 that on Earth.)
Physics
1 answer:
Liula [17]4 years ago
5 0
The distance an object falls from rest through gravity is 
                        D  =  (1/2) (g) (t²) 
           Distance  =  (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected 
if  ' D ' doesn't change but ' g ' does. 
So I'm going to start by rearranging the equation
to solve for ' t '.                                                      D  =  (1/2) (g) (t²)

Multiply each side by  2 :         2 D  =            g    t²  

Divide each side by ' g ' :      2 D/g =                  t² 

Square root each side:        t = √ (2D/g)

Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:

  -- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'

                                             and smaller 'g' ==> longer 't' .-- 

They don't change by the same factor, because  1/g  is inside the square root.  So 't' changes the same amount as  √1/g  does.

Gravity on the surface of the moon is roughly  1/6  the value of gravity on the surface of the Earth.

So we expect ' t ' to increase by  √6  =  2.45 times.

It would take the same bottle  (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.
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Answer :

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Explanation:

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Arte-miy333 [17]
The correct answer is "an attractive force" between the wires.

Let's see why. Assume we have wire A on the left and wire B on the right, and that the current in both wires go upward.  First, let's find the direction of the magnetic field produced by wire A at wire B: by using the right-hand rule, we see that since the current (the thumb) goes upward, the magnetic field (given by the other fingers) at wire B is directed inside the paper.
Then we can apply again the right-hand rule to see what is the force on wire B. The index gives the direction of the current (upward), the middle finger the direction of the magnetic field (inside the paper), and the thumb gives the direction of the force: to the left, so toward wire A. This means the force is attractive. (you can re-do the procedure on wire A, and you will find the force on wire A is directed toward wire B)
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A storm is moving east towards your house at an average speed of 35 km/hr. If the storm is currently 80 kilometers from your hou
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Forces in a Three-Charge System Coulomb's law for the magnitude of the force F between two particles with charges Q and Q′ separ
Dmitriy789 [7]

Answer:

Explanation:

Given the charges.

Q1=-17.5nC. Negative charge

Q2=32.5nC. Positive charge

Q3=55nC. Positive charge

Q1 is at a distance of -1.68m on the x-axis

Q2 is at the origin i.e at 0m

Q3 is at between Q1 and Q2 at -1.085m on the x-axis

It shows that,

Q1 is at -1.085+1.68 =0.595m from Q3

Also, Q2 is at 1.085m from Q3.

K=9×10^9Nm²/C²

We need to find the net force on Q3

Then we need F13 and F23

Firstly F13

Between Q1 and Q3

There will be attraction i.e, Q3 will move to the negative direction of the x axis, then F13 will be in negative direction

So,

F13=kQ1Q3/r²

F13=9E9×17.5E-9×55E-9/0.595²

F13=2.45×10^-5N

In vector form

F13=—2.45×10^-5N i

Now, we need F23,

This will the force of repulsion because they are both positive charge, the the charge Q3 will move to the negative direction of x axis, since Q2 is at the origin and Q3 is at negative x axis. So, F23 will be negative

F23=kQ2Q3/r²

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In vector form

F23=—1.367×10^-5N i

Then the net force is given as

Fnet = F13+F23

Fnet=—2.45×10^-5Ni—1.367×10^-5Ni

Fnet=—3.82×10^-5N i

Magnitude for the Fnet is

Fnet=3.82×10^-5N.

And the direction

θ= arctan(y/x).

y=0 and x=3.82×10-5

θ= arctan(0/-3.82E-5)

θ=arctan(-0)

θ= 0. in the negative direction, i.e 180°.

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