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Fynjy0 [20]
2 years ago
5

The gravitational force between Pluto and Charon is 3.61 × 1018 N. Pluto has a mass of 1.3 × 1022 kg, which is only slightly gre

ater than Charon’s mass of 1.6 × 1021 kg. How far apart are Pluto and Charon? 2.0 × 107 m 2.4 × 1012 m 3.8 × 1014 m 5.8 × 1024 m
Physics
2 answers:
raketka [301]2 years ago
4 0

<u>Answer:</u>

<em>The distance between Pluto and Charon is  1.96 \times 10^7  m</em>

<u>Explanation:</u>

Force of gravitation between two objects F_g= \frac{G(M_1 M_2 )}{r^2}  

Where M_1  \ and \ M_2  are the masses of the objects,r is the distance between the objects

G is the universal gravitational constant=6.67 \times 10^-^1^1 m^3/kg s^2

Here mass of pluto = 1.3 \times 10^2^2  kg

mass of charon = 1.6 \times 10^2^1 kg

Force of gravitation F_g=3.61 \times 10^1^8  N

F_g= \frac {G(M_1 M_2 )}{r^2}

r^2= \frac {G(M_1 M_2 )}{F_g }

r= \sqrt \frac{(G(M_1 M_2 )}{F_g}

=\sqrt \frac {((6.674\times 10^-^1^1) \times 1.3 \times 10^2^2 \times 1.6\times 10^2^1)}{(3.61 \times 10^1^8 )}

=\sqrt \frac {(13.88*10^32)}{(3.61*10^18)}

=1.96 \times10^7  m

chubhunter [2.5K]2 years ago
4 0

Answer:

The answer is A on edgen.

Explanation:

A. 2.0 × 10^7 m

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The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. what percentage of the mechanical energy of th
Zanzabum

E = ½KA^2 is the mechanical energy of any oscillator.  It is the sum of elastic potential energy and kinetic energy.  When amplitude A decreases by 3%, then

(E2-E1)/E1 = {½K(A2^2/A1^2) }/ ½K(A1^2)

= {(A2^2 – A1^2) / (A1^2)}

= 97^2 – 100^2/100^2

= 5.91% of the mechanical energy is lost each cycle.

3 0
1 year ago
Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una
EleoNora [17]

Answer:

a)  f = 615.2 Hz      b)  f = 307.6 Hz

Explanation:

The speed in a wave on a string is

         v = √ T / μ

also the speed a wave must meet the relationship

          v = λ f

           

Let's use these expressions in our problem, for the initial conditions

            v = √ T₀ /μ

             √ (T₀/ μ) = λ₀ f₀

now it indicates that the tension is doubled

         T = 2T₀

          √ (T /μ) = λ f

          √( 2To /μ) = λ f

         √2  √ T₀ /μ = λ f

we substitute

         √2 (λ₀ f₀) = λ f

if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change

           λ₀ = λ

           f = f₀ √2

           f = 435 √ 2

           f = 615.2 Hz

b) The tension is cut in half

         T = T₀ / 2

         √ (T₀ / 2muy) =  f = λ f

          √ (T₀ / μ)  1 /√2 = λ f

           fo / √2 = f

           f = 435 / √2

           f = 307.6 Hz

Traslate

La velocidad en una onda en una cuerda es

         v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

          v= λ f  

           

Usemos estas expresión en nuestro problema, para las condiciones iniciales

            v= √ To/μ

             √ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

         T = 2T₀

          √ ( T/μ) = λf

          √ ) 2T₀/μ = λ f

         √ 2 √ T₀/μ = λ f

         

substituimos  

         √2    ( λ₀ f₀)  =  λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

                 λ₀ =  λ

           f = f₀ √2

           f = 435 √2

           f = 615,2 Hz

b)  La tension se reduce a la mitad

         T = T₀/2    

         RA ( T₀/2μ)  =  λ  f

          Ra(T₀/μ) 1/ra 2  =  λ f

           fo /√ 2 = f

           f = 435/√2

           f = 307,6 Hz

5 0
2 years ago
Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic​
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The volume of the room is the product of its dimensions:

10\times 12 \times 4 = 480\text{ m}^3

Now, from the equation

d=\dfrac{m}{V}

where d is the density, m is the mass and V is the volume, we deduce

m=dV

So, multiply the density and the volume to get the mass of air in the room.

7 0
2 years ago
A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0
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Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

vsin(\alpha) + gt = 0

t = \frac{-vsin(\alpha)}{g}

So the vertical distance it travels within time t is

y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}

y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}

y = \frac{-v^2sin^2(\alpha)}{2g}

Similarly the horizontal distance it travels within time t is:

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x = \frac{-v^2sin(2\alpha)}{2g}

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So y = 57.54sin^2(\alpha)

x = 57.54sin(2\alpha)

From here we can plug-in the angles values of 30, 45 and 60 degrees

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

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