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dusya [7]
3 years ago
14

If every action has an equal and opposite reaction, then when you jump, and gravity pulls you back down, it also pulls the Earth

up to you. Why can’t you see this motion?
Physics
1 answer:
LenKa [72]3 years ago
8 0

When two objects are gravitationally attracted, the forces

on both of them are equal. As the objects move straight

toward each other, then, just as we'd expect, the object

with the greater mass has the smaller acceleration and

moves a smaller distance by the time they collide, while

the object with a smaller mass has the greater acceleration

and moves a greater distance by the time they collide.


The Earth's mass is roughly approximately something like 70,000,000,000,000,000,000,000 times as much as YOUR mass,

so the difference in acceleration and distance moved before

the collision is really substantial.

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Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
An incident ray that passes through the vertex of a convex lens:
WINSTONCH [101]
The answer is refracts parallel to the axis of the lens
7 0
3 years ago
Rajani had bought a new bottle of pickle from the market. She tried to open the
dem82 [27]

Answer:

The pickle bottle cap on dipping it in hot water expanded.

Explanation: Hope it helps you:))))

Have a good day

4 0
2 years ago
From Doppler shifts of the spectral lines in the light coming from the east and west edges of the Sun, astronomers find that the
Gala2k [10]

Answer:

T= 37 day

Explanation:

To solve this exercise we will use the definition of angular velocity as the angular distance, which for a full period is 2pi between time.

    w = T / t

The relationship between angular and linear velocity is

    v = w r

    w = v / r

We substitute everything in the first equation

    v / r = 2π / t

    t = 2π r / v

Let's reduce to the SI system

    V = 2 km / s (1000m / 1km) = 2 10³ m / s

    r= R = 6.96 10⁸ m

Let's calculate

    t = 2π 6.96 10⁸/2 10³

    t = 3.2 10⁶ s

    T = t = 3.2 10⁶ s ( 1h/3600s) (1 day/24 h)

    T= 37 day

3 0
3 years ago
An aircraft is at a standstill. It accelerates to 140kts in 28 seconds. The aircraft weighs 28,000 lbs. How many feet of runway
stiv31 [10]

Answer:

runway use is 3307.8 feet

Explanation:

given data

velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s

time = 28 seconds

weight = 28000 lbs

to find out

How many feet of runway was used

solution

we will use here first equation of motion for find acceleration

v = u + at     ..............1

here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time

put here value in equation 1

72.016 = 0 + a(28)

a = 2.572 m/s²

and

now apply third equation of motion

s = ut + 0.5×a×t²    .......................2

here s is distance and u is initial speed and t is time and a is acceleration

put here all value in equation 2

s = 0 + 0.5×2.572×28²  

s = 1008.24 m = 3307.8 ft

so  runway use is 3307.8 feet

5 0
4 years ago
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