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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.
You might be interested in
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
2 years ago
somewhere between the earth and the moon is a point where the gravitational attraction of the earth is canceled by the gravitati
mote1985 [20]
<span>It's pretty easy problem once you set it up.

Earth------------P--------------Moon

"P" is where the gravitational forces from both bodies are acting equally on a mass m

Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon

You are correct to use that equation. If the gravitational forces are equal then

GMearth*m/Rep² = Gm*Mmoon/Rpm²

Mearth/Mmoon = Rep² / Rpm²

Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be

Rpm = Rem - Rep

Mearth / Mmoon = Rep² / (Rem - Rep)²

Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²

Everything is done now. The most complicated part now is the algebra, so bear with me as we solve for Rep. I may skip some obvious or too-long-to-type steps.

81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0

We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem

Obviously, point P cannot be 9/8 of the way to the moon because it'll be beyond the moon. Therefore, the logical answer would be 9/10 the way to the moon or B.

Edit: The great thing about this idealized 2-body problem, James, is that it is disguised as a problem where you need to know a lot of values but in reality, a lot of them cancel out once you do the math. Funny thing is, I never saw this problem in physics during Freshman year. I saw it orbital mechanics in my junior year in Aerospace Engineering. </span> sylent_reality · 8 years ago
8 0
2 years ago
Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68
den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, y_0=1500\ m

Initial horizontal velocity is, u_x=688\ m/s

Initial vertical velocity is, u_y=0(\textrm{Horizontal velocity only initially)}

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is S=y-y_0=0-1500=-1500\ m

Acceleration in the vertical direction is due to gravity, g=-9.8\ m/s^2

Therefore, the displacement of the bomb is given as:

S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0
2 years ago
The A-string (440 HzHz) on a piano is 38.9 cmcm long and is clamped tightly at both ends. If the string tension is 667N, what's
Mice21 [21]

Answer:

Mass, m = 2.2 kg                                

Explanation:

It is given that,

Frequency of the piano, f = 440 Hz

Length of the piano, L = 38.9 cm = 0.389 m

Tension in the spring, T = 667 N

The frequency in the spring is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{L} is the linear mass density

On rearranging, we get the value of m as follows :

m=\dfrac{T}{4Lf^2}

m=\dfrac{667}{4\times 0.389\times (440)^2}

m = 0.0022 kg

or

m = 2.2 grams

So, the mass of the object is 2.2 grams. Hence, this is the required solution.

3 0
2 years ago
Read 2 more answers
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur
GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  \mu mg  =  \frac{mv^2}{r}

    where \mu is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0
2 years ago
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