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VARVARA [1.3K]
3 years ago
13

Question: Why do both the moon and the earth pull on the spaceship the entire time? Why must you be closer to the moon before th

e pull of the earth and moon are equal?
Physics
1 answer:
Ira Lisetskai [31]3 years ago
8 0
-- There's no limit to the distance of gravitational forces. 
There's gravitational force between Pluto and the lint in your
pocket ... not much, but it's there, and it can be calculated.

So there's ALWAYS gravitational force between the Earth and the
spaceship, AND ALSO between the Moon and the spaceship. 
Even before it's ever launched !

-- The Earth has about 80 times as much mass as the Moon has,
so you have to be much closer to the Moon before the gravitational
forces in each direction are equal.
You might be interested in
I need help on (a)<br> I don't know what equation to use?
Alchen [17]

Impulse = (force) x (length of time the force lasts)

I see where you doodled  (60)(40)  over on the side, and you'll be delighted
to know that you're on the right track !

Here's the mind-blower, which I'll bet you never thought of:
On a force-time graph, impulse (also change in momentum)
is just  the <em>area that's added under the graph during some time</em> !

From zero to 60, the impulse is just the area of that right triangle
under the graph.  The base of the triangle is  60 seconds.  The
height of the triangle is  40N .  The area of the triangle is not
the whole (base x height), but only <em><u>1/2 </u></em>(base x height).

  1/2 (base x height) = 1/2 (60s x 40N) = <u>1,200 newton-seconds</u>

<u>That's</u> the impulse during the first 60 seconds.  It's also the change in
the car's momentum during the first 60 seconds. 

Momentum = (mass) x (speed)

If the car wasn't moving at all when the graph began, then its momentum is  1,200 newton-sec after 60 seconds.  Through the convenience of the SI system of units, 1,200 newton-sec is exactly the same thing as 1,200 kg-m/s .  The car's mass is 3 kg, so after 60 sec, you can write

    Momentum = M x V = (3 kg) x (speed) = 1,200 kg-m/s

and the car's speed falls right out of that. 

From 60to 120 sec, the change in momentum is the added area of that
extra right triangle on top ... it's 60sec wide and only 20N high.  Calculate
its area, that's the additional impulse in the 2nd minute,  which is also the
increase in momentum, and that'll give you the change in speed.


8 0
4 years ago
A leaky 10-kg bucket is lifted from the ground to a height of 14 m at a constant speed with a rope that weighs 0.5 kg/m. Initial
Lina20 [59]

solution:

Weight of bucket = 10kg

Length or distance =14m

Weight of rope=0.5kg/m

At any point x of the rope,

=(0.5)(14-x)

=(7-0.5x)

Since the water finishes draining at 14m level and total weight of water is 42kg

Total mass=(7-0.5x)+(42-3x)+10=(59-3.5x)kg

Force=(9.8)(59-3.5x)

work w =\lim_{n \to \infty }\sum_{i \to 1}^{n}(9.8)(59-3.5x)\Delta x\\=\int_{0}^{14}(9.8)(59-3.5x)dx\\=9.8\int_{0}^{14}(59-3.5x)dx\\9.8((59x-\frac{3.5x^2}{2})){_{0}}^{14}\\9.8(59(14)-\frac{3.5(14)^2}{2})\\=4733.4\\therefore,\\W=4733.4J\approx 4733J

5 0
3 years ago
1. A plane starts from rest and aceelerates in a
larisa86 [58]

s=600 m

t=12 s

s=0.5*a*t² (initial speed V0=0)

a=(2*s)/t²

a=(2*600)/12²

a≈8.33 m/s²

L= s(t2=12s)-s(t1=11s) -> (distance during the twelfth second)

L=0.5*a*(t2²-t1²)

L=0.5*((2*s)/t²)*(t2²-t1²)

L=0.5*((2*600)/12²)*(12²-11²)

L ≈ 95.83 m

6 0
3 years ago
If you ride your bike at an average speed of 2 km/h and need to travel a total distance of 20 km, how long will it take you to r
love history [14]

Answer:

Time taken to reach your destination will be 10hours

Explanation:

Recall the formula for Speed;

speed=Total distance/Total time taken

Speed=2km/h

Total distance=20km

Time taken=x

let x be the unknown time taken

Input each values into the formula;

2=20/x

Making x subject of the equation

x=20/2

x=10

Total time taken =10hours.

3 0
3 years ago
Read 2 more answers
Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, w
olga nikolaevna [1]

To solve this problem, apply the concepts related to the calculation of the work performed according to the temperature change (in an ideal Carnot cycle), for which you have to:

W = \int\limit_{T_c}^{T_H} C (1-\frac{T_H}{T})

Where,

C = Heat capacity of the Brick

T_C= Cold Temperature

T_H = Hot Temperature

Integrating,

W = C (T_H-T_C)- T_H C ln (\frac{T_H}{T_C})

Our values are given as

T_H= 300K

T_C = 150K

Replacing,

W = (1) (300-150)-300(1)ln(2)

W = 150-300ln2

W = -57.94kJ \approx 58kJ

Therefore the work perfomed by this ideal carnot engine is 58kJ

5 0
4 years ago
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