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Debora [2.8K]
2 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]2 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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A CCD has a greatest possible pixel value of 4095. what is the bit level of this CCD?
Shtirlitz [24]
<span>Bit level for a CCD (Charged coupled device) with a greatest possible pixel value of 4095:The relationship between the bit level and pixel value is given as:pixel value = 2^bit level.Most charged coupled devices (CCDs) have 8-bit, 16-bit, 32-bit levels.Using simple mathematics, we can see that 2^12 = 4096.Since the maximum number of pixels is 4095, the bit level is 12., i.e. the CCD has 12-bit level.</span>
8 0
2 years ago
1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational
sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = \frac{m v^{2} }{r}

==> 1.1 × 10^{19} = (9.4 × 10^{21} × v^{2} ) ÷ 1.5 × 10^{8}

==> 1.1 × 10^{19} = v^{2} × 6.26×10^{13}

==> v^{2} =  1.1 × 10^{19} ÷ 6.26×10^{13}

==> v^{2} = 0.17571885 × 10^{6}

==> v= 0.419188323 × 10^{3} m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×10^{8} ÷ 0.419188323 × 10^{3}

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0
2 years ago
Read 2 more answers
Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc
Lilit [14]

Answer:

Explanation:

Image of distant object will be made at far point or at 52.5 so

object distance u = infinity

image distance v = - 52.5 cm

focal length required = f

Lens formula

1 / v - 1 / u = 1 / f

1 /  - 52.5 - 0 = 1 / f

f =  -52.5 cm

= -.525 m

Power P = 1 / f = -  1 / .525

= -  1.90

now , for eye with glass we shall find new near point .

v = ?

u = - 17.2 cm

f = -  52.5 cm

1 / v - 1 / u = 1 / f

  1 / v + 1 / 17.2 = -  1 / 52.5

1 / v  = - 1 / 17.2 -    1 / 52.5

= - .05813 -  .019

= - .07713

u = - 12.96 cm

so new near point will be 12.96 cm

5 0
2 years ago
What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u
Aleksandr-060686 [28]

Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 ×  10^{14} Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

E=h\nu_{o}    ..................1

here E is the energy of electron per atom E=\frac{x}{N}  and h is plank constant i.e. 6.626\times10^{-34} Js  and x is  binding energy

and here N is the Avogadro constant = 6.023\times10^{23}

so E will \frac{x}{6.023\times10^{23}}  

E = 3.19\times10^{-19}  J  

so put value in equation 1 we get

\frac{x}{6.023\times10^{23}} = 2.50 ×  10^{14} × 6.626\times10^{-34} Js  

solve it we get

x = 99770.99

so  binding energy is 99771 J/mol

4 0
2 years ago
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25
antiseptic1488 [7]

we are given in the problem the following dimensions or specifications 
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7 

The formula that is applicable from physics is 
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes 
7 0
1 year ago
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