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dusya [7]
3 years ago
6

40 Points!!!!!!!!!!!!!

Physics
1 answer:
Firlakuza [10]3 years ago
3 0
The answer is A for sure
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Good night all of you​
Helen [10]

Answer: Good night

Explanation:

3 0
3 years ago
Read 2 more answers
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
You have a friend who reports that he falls asleep easily around 11 PM but then awakens for about an hour most nights around two
Solnce55 [7]

Answer:

can I have more details please, Thanks!

Explanation:

5 0
3 years ago
During the first 40 s a truck traveled at a constant speed of 15 m/s. Find the
kotykmax [81]

Answer:

600 m

Explanation:

Distance = speed × time

= (15)(40)

= <u>6</u><u>0</u><u>0</u><u> </u><u>m</u>

6 0
3 years ago
Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

Frequency of Source,

f_{s}=500\ Hz

To Find:

Frequency of Observer,

f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source v_{s} and observer v_{o}, the original frequency of the sound waves f_{s} and the observed frequency of the sound waves f_{o},

The Equation is

f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
3 years ago
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