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Stels [109]
3 years ago
5

What force holds the earth-moon system together?

Physics
2 answers:
Verdich [7]3 years ago
7 0
Gravity holds the system together
valentinak56 [21]3 years ago
7 0
The force of gravity, attracting the Earth and the Moon to each other,
keeps the two bodies in orbit around their common center of mass.

(Gravity also keeps the common center of Earth-mass and Moon-mass
in orbit around the sun.)
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A Russian athlete lifts 300kg to a height of 2.5 m in 2 seconds. What is the work done? And what power is exerted?
snow_tiger [21]
We want to find the work done and power exerted, let’s start with work first.

We know that the equation for work is: W = F * D. We need to find the force which we can find by using: F = M * A.

Mass: 300kg
Acceleration (using equation from photo): 1.25 m/s^2

(The equation says x but can be used with y values)

If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.

To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.

Force = 300 kg * 1.25 m/s^2 = 375 Newtons

So, work = 500 N * 2.5 m = 1000 Joules

Power: W/t

So, Power = 1000 J / 2 seconds = 500 Watts

Hope this helps!

5 0
3 years ago
If two charges attract each other, the two charges must be
padilas [110]

Answer:

be negative

Explanation:

8 0
4 years ago
Read 2 more answers
Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa.  Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

  • A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.
  • v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
  • D2: 0.04 m
  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

8 0
3 years ago
A player kicks a soccer ball from ground level and sons at flying at an angle of 30° at a speed of 26MS how far did the ball tra
RSB [31]

Answer:

Explanation:

60 meters is he answer for this question

7 0
4 years ago
A 65.0 kg diver is 4.90 m above the water, falling at speed of 6.40 m/s. Calculate her kinetic energy as she hits the water. (Ne
mojhsa [17]

Answer:

4452.5 J.

Explanation:

The diver have both kinetic and potential energy.

Ek = 1/2mv² ................. Equation 1

Where Ek = Kinetic Energy of the diver, m = mass of the diver, v = velocity of the diver.

Given: m = 65 kg, v = 6.4 m/s.

Substitute into equation 1

Ek = 1/2(65)(6.4²)

Ek = 1331.2 J.

Also,

Ep = mgh ............................ Equation 2

Where Ep =  Potential energy of the diver when its above the water, h = height of the diver above the water, g = acceleration due to gravity.

Given: m = 65 kg, h = 4.9 m, g = 9.8 m/s²

Substitute into equation 2.

Ep = 65(4.9)(9.8)

Ep = 3121.3 J.

Note: When she hits the water, the potential energy is converted to kinetic energy.

E = Ek+Ep

Where E = Kinetic energy of the diver when she hits the water.

E = 1331.2+3121.3

E = 4452.5 J.

3 0
3 years ago
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