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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 woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow
Ymorist [56]

Answer:

Explanation:

Given

Force P is acting upward

C is vertical contact Force

W is the weight of the crate

As P is unable to move the Block therefore Normal reaction keeps on acting on block

thus we can say that

P-W+C=0

P=W-C

                   

7 0
2 years ago
Read 2 more answers
The wind blows a jay bird south with a force of 300 Newtons. The
adoni [48]

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The  jay bird flies north, against the wind, with a force of 100  newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

F=\sqrt{300^2+100^2} \\\\F=316.22\ N

Hence, the net force on the jay bird is 316.22 N.

6 0
2 years ago
A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle
alexira [117]
<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>
  • According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.
  • Therefore; Sin i/Sin r = η

In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

       = 0.3759

r = Sin^-1 0.3759

   = 22.08

   <u>≈ 22°</u>

3 0
2 years ago
Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B
gregori [183]

Answer:

The force P required  is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

tan \ \alpha = \dfrac{CD}{BD}

where;

BD = AB-AD

Then;

tan \ \alpha = \dfrac{CD}{AB-AD}

\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )

replacing their respective values, where;

CD = 2 sin 30° m,  AB = 2m and AD = 2 cos 30° m

\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )

\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )

\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )

\alpha = tan^{-1} \bigg(3.73\bigg )

\alpha \simeq 75^0

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

T_{BC} = P

Using the force equilibrium expression along the horizontal direction.

\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0

replacing the value of \alpha \simeq 75^0

-T_{AC} \  cos 30^0 + P cos 75^0  = 0

P \ cos \ 75^0 = T_{AC} \ cos \ 30^0

P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - -  (1)

Along the vertical direction, the force equilibrium equation can be expressed as:

\sum F_y =0

-W + P \ sin \alpha + T_{AC} \ sin \ 30^0  = 0

W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0

replacing \alpha \simeq 75^0 and P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

Also, replacing W for (200 × 9.81) N

200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0

1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)

1962= T_{AC} \ (0.8660\times 3.732 + 0.5)

1962= T_{AC} \ (3.231912 + 0.5)

1962= T_{AC} \ (3.731912)

T_{AC}  = \dfrac{1962}{ \ (3.731912)}

T_{AC}  = 525.736 \ N

From P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}

P =\dfrac{ 525.736 \times0.866}{0.2588}

P = 1759.22 N

Thus, the force P required  is 1759.22 N

6 0
1 year ago
When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.
gayaneshka [121]

Answer:

Cis, Trans.

Explanation:

Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.

Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.

5 0
2 years ago
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