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Alexeev081 [22]
3 years ago
13

Calculate the position of the center of mass of the following pairs of objects. Use a coordinate system where 0 is at the center

of the more massive object. Give your answer not in meters but as a fraction of the radius as requested. You can find the data you need in the inside of the back cover of your textbook. (a) The Earth and the Moon. Give the answer as a fraction of the earth's radius.
Physics
1 answer:
vlada-n [284]3 years ago
8 0

Answer:

the center of mass is  Rcm = 0.725 R Earth

Explanation:

The concept of center of mass is of great importance in all physics, since on this all external forces of the system can be considered applied, in the collision analysis it gives an ideal reference system for resolution

The equation for the center of mass is

        Rcm = ∑ mi ri / Mt

With Rcm and ri being vector quantities, "m" is body mass and MT total mass

If we write this equation for our system we have

       Rcm = (M1 r1 + M2 r2) / (M1 + M2)

M1 and M2 are the mass of the Earth and the Moon respectively, r1 and r2 are the distances measured from our reference system. As they indicate that we place the most massive reference system we place it on planet Earth.  The data taken a textbook.

       

        M1 = 5.98 10²⁴  Kg

        M2 = 7.36 10²² Kg

       Distance from Earth to the moon on average 3.84 10⁸ m

How the coordinate system is placed on Earth

 

        r1 = 0

        r2 = 3.84 10 8 m

      Rcm = 0 + 7.36 10²²  3.84 10⁸ / (5.98 10²⁴ + 7.36 10²²)

R cm = 2.826 10³¹ / (598 10²² + 7.36 10²²)

Rcm = 2.826 10³¹ / 611.34 10²²

R cm = 4.62 10⁶  m

we are asked to give this distance as a fraction of the radius of the Earth, we divide the two quantities

      R earth = 6.37 10⁶ m

     Fraction = R cm / R Earth

     Fraction = 4.62 10⁶ / 6.37 10⁶

     Fraction = 0.725

In summary the center of mass is

     Rcm = 0.725 R Earth

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vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2 (1)

where

p_1 = 7340 N/m^2 is the pressure in the pipe

p_2 = 5505 N/m^2 is the pressure in the constricted section

\rho = 821 kg/m^3 is the density of the oil

v_1 is the velocity of the oil in the pipe

v_2 is the velocity of the oil in the constricted section

Also, according to the continuity equation,

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the pipe, with

r_1 = \frac{1.03}{2}=0.515 m is the radius

A_2 = \pi r_2^2 is the cross-sectional area of the constricted section, with

r_2=\frac{0.618}{2}=0.309 m is the radius

So the equation becomes

r_1^2 v_1 = r_2^2 v_2

So we can write

v_2=\frac{r_1^2}{r_2^2}v_1

Substituting into eq.(1),

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2

And solving the equation for v_1:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s

And the velocity in the constricted section is

v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s

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7 0
3 years ago
An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place
anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

mu=(M+m)v

v=\frac{m}{M+m}\times u

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

\frac{(M+m)v^2}{2}=(M+m)gh

here h=L(1-\cos \theta ) from diagram

therefore

v=\sqrt{2gL(1-\cos \theta )}

initial velocity in terms of v

u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}

For first case \theta =6.8^{\circ}

u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}

for second case \theta =11.4^{\circ}

u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}

Therefore \frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}

\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}

\frac{u_1}{u_2}=1.084

i.e.\frac{v_1}{v_2}=1.084

4 0
3 years ago
Why is physics used to study stars?
goblinko [34]

Answer:

stars are made of matter

3 0
2 years ago
As you lift an 88 N box straight upward you produce a power of 72 W. What is the speed of the box?
jeka57 [31]

F = force applied to lift the box = weight of the box = 88 N

P = power produced while lifting the box upward = 72 Watt

v = speed of the box = ?

Speed of the box is given as

v = \frac{P}{F}

inserting the values

v =  \frac{72}{88}

v = 0.82 ms⁻¹

6 0
3 years ago
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An electron is projected with an initial speed Vo = 5.35x10^6 m/s into the uniform field between the parallel plates. The direct
mart [117]

Answer:

 E = 3.04 10⁻⁵ N / C

Explanation:

In this problem we can use the kinematics to find how long it takes the electron to travel the plates

Let's start by reducing the magnitudes to the SI system

              vₓ = 5.35 10⁶ m / s

              x = 2 cm = 2 10⁻² m

              y = 1 cm = 1 10⁻² m

             x = vₓ t

             t = x / vₓ

             t = 2 10⁻² / 5.35 10⁶

             t = 3,738 10⁻⁹ s

This time is also the time it takes for vertical movement to go from the center to the plate, let's look for acceleration with Newton's second law

              F = m a

              a = F / m = e E / m

              y = v_{oy} + ½ a t²

              v_{oy} = 0

We replace

              y = ½ e / m E t²

              E = 2 y m / e t²

Let's calculate

            E = 2  1 10⁻²  9.1 10⁻³¹ / (1.6 10⁻¹⁹  3,738 10⁻⁹)

            E = 18.2 10⁻³³ / 5.98 10⁻²⁸

            E = 3.04 10⁻⁵ N / C

5 0
2 years ago
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