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Nady [450]
3 years ago
5

Does the water stay at the same speed all the way down the mountain from the reservoir ? If not, why not? please answer

Physics
1 answer:
photoshop1234 [79]3 years ago
3 0

Answer:

No

Explanation:

Because the lower part of the water is at lower speed since it is where most aquatic animals live

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How did Carl Rogers view personality?
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Carl Rogers was an influential humanistic psychologist who developed a personality theory that emphasized the importance of the self-actualizing tendency in shaping human personalities. ... Human beings develop an ideal self and a real self based on the conditional status of positive regard

Explanation:

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Help??? Phases of the moon. Am I right
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Yes. You are correct. Great job!

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4 years ago
Un the way to the moon, the Apollo astro-
kherson [118]

Answer:

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

6 0
3 years ago
En una báscula hidráulica colocamos una persona de 75 kg sobre un émbolo y un camión de 7200 kg sobre una plataforma de 5 m de l
TEA [102]

...........................

4 0
4 years ago
A toy spacecraft is launched directly upward. When the toy reaches its highest point, a spring is released and the toy splits in
Murljashka [212]

Answer:

A

Explanation:

Momentum conservation will cause 0.08kg to move to the west (opposite of 0.02 kg).

and because both are at the same height above the ground, they will take the same time to reach the ground.

The speed of 0.08kg will be less than 0.02 kg, let v be the speed of 0..02kg, then speed of 0.08kg V is

0.02v - (0.08)V = 0

V = 0.02 v/ 0.08 = v/4

The speed of 0.08 kg = v/4

The speed of 0.08 kg is less than 0.02kg.

So 0.02kg strikes the ground farther from the launch point than does the 0.08 kg

8 0
3 years ago
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