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Mazyrski [523]
3 years ago
8

Pluto’s diameter is approximately 2370 km, and the diameter of its satellite Charon is 1250 km. Although the distance varies, th

ey are often about 19,700 km apart, center to center. Assuming that both Pluto and Charon have the same composition and hence the same average density, find the location of the center of mass of this system relative to the center of Pluto.
Physics
1 answer:
MissTica3 years ago
5 0

Answer:

r_{cm} = 2520.5 km

Explanation:

As we know that mass is the product of volume and density

so we will have

M = \rho V

here we have

M = \rho(\frac{4}{3}\pi r^3)

so we will have

\frac{M_p}{M_c} = (\frac{r_p}{r_c})^3

so we will have

\frac{M_p}{M_c} = (\frac{2370}{1250})^3

M_p = 6.81 M_c

now let the position of Pluto is at origin so we have

r_{cm} = \frac{M_p (0) + M_c(19700)}{M_p + M_c}

r_{cm} = \frac{19700}{\frac{M_p}{M_c} + 1}

r_{cm} = \frac{19700}{6.81 + 1}

r_{cm} = 2520.5 km

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4 0
3 years ago
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______ uses radioactive materials to create contrast in the body and help form images of the structure and function of organs.
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Answer:

Nuclear medicine or PET scanning (Positron Emission Tomography)

Explanation:

Not super sure what your question is looking for but I think it can be either. PET is a technology used in the nuclear medicine field, and nuclear medicine is a broad field of using the technology the question described.

8 0
2 years ago
A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an e
Gala2k [10]

Explanation:

Given that,

Charge acting on the object, q=-4\ nC=-4\times 10^{-9}\ C

Force acting on the object, F=19\ nC=19\times 10^{-9}\ C (in downward direction)

(a) The electric force acting in the electric field is given by :

F=qE

E is the electric field

E=\dfrac{F}{q}

E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, q=1.6\times 10^{-19}\ C

The force acting on the proton is :

F=qE

F=1.6\times 10^{-19}\times 4.75

F=7.6\times 10^{-19}\ N

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

8 0
3 years ago
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
Strike441 [17]

Answer:

E=3.5(8.98*10^{6}x-2.69*10^{15}t)

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)

Explanation:

The electric field equation of a electromagnetic wave is given by:

E=E_{max}(kx-\omega t) (1)

  • E(max) is the maximun value of E, it means the amplitude of the wave.
  • k is the wave number
  • ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

k=\frac{2\pi}{\lambda}            

k=8.98*10^{6} [rad/m]      

And the relation between λ and f is:                

c=\lambda f

f=\frac{c}{\lambda}

f=\frac{3*10^{8}}{700*10^{-9}}

f=4.28*10^{14}

The angular frequency equation is:

\omega=2\pi f

\omega=2\pi*4.28*10^{14}

\omega=2.69*10^{15} [rad/s]

Therefore, the E equation, suing (1), will be:

E=3.5(8.98*10^{6}x-2.69*10^{15}t) (2)

For the magnetic field we have the next equation:

B=B_{max}(kx-\omega t) (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

E_{max}=cB_{max}

B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}

B_{max}=1.17*10^{-8}T

Putting this in (3), finally we will have:

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t) (4)

I hope it helps you!

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3 years ago
Velocity is:
sleet_krkn [62]

Answer:

d

Explanation:

Solution:-

- The Quantity of theory of money states:

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Where,

           M = Money supply

           V = Velocity of money exchange

           P = The price level

           Y = Real GDP

- By re-arranging the formula and solving for "V" we have:

                     V = P*Y / M

- The expression on right hand side increases if exchange of dollars increases.

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