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liberstina [14]
2 years ago
14

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance

of 2.7×1011 m from the center of the sun, what is its speed when at a distance of 4.8×1010 m?
Physics
1 answer:
xz_007 [3.2K]2 years ago
3 0

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

v₂ = 7.6 x 10⁴ m/s

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A group of 25 particles have the following speeds:
svp [43]

Answer:

Part a: The average speed is 24.12 m/s

Part b: The rms speed is 25.55 m/s

Part c: The most probable speed is 17 m/s.

Explanation:

Part a

Average Speed

<em>Average speed is given as </em>

<em />v_{avg}=\frac{\sum}{n}

v_{avg}=\frac{(2 \times 11) +(7 \times 17)+(4 \times 19)+(3 \times 27) +(6 \times 32) +(1 \times 33)+(2 \times 40)  }{25}\\v_{avg}=\frac{22+119+76+81+192+33+80}{25}\\v_{avg}=\frac{603}{25}\\v_{avg}=24.12 m/s

So the average speed is 24.12 m/s

Part b

RMS Speed

v_{rms}=\frac{\sum v^2}{n}

v_{rms}=\sqrt{\frac{(2 \times 11^2) +(7 \times 17^2)+(4 \times 19^2)+(3 \times 27^2) +(6 \times 32^2) +(1 \times 33^2)+(2 \times 40^2)  }{25}} \\v_{rms}=\sqrt{\frac{242+2023+1444+2187+6144+1089+3200}{25}}\\\\v_{rms}=\sqrt{\frac{16329}{25}}\\v_{rms}=\sqrt{653.16}\\v_{rms}=25.55 m/s

So the rms speed is 25.55 m/s

Part c

Most Probable Speed

As 7 particles have speed of 17 m/s i.e. 7 is the highest frequency so 17 m/s is the most probable speed.

3 0
3 years ago
A train whistle is 580 Hz when stationary. It is moving away from you at 18.8 m/s. What frequency do you hear?
Natalija [7]

Answer

613.63Hz

Explanation:

Fobs = (V/V-Vsource)Force

V=speed of sound

Fobs= (343/343-18.8)580

Fobs= 613.63Hz

Fobs = frequency observed

brainly.com/question/15339151

8 0
2 years ago
Two runners ran side by side, each holding one end of a horizontal pole. How would this affect the direction of the runners? Exp
zhuklara [117]

Answer:

They will run parallel to each other as the none of a straight pole cannot be bent in such a way where one side can turn without the other turning.

6 0
2 years ago
ANSWER THIS FOR BRAINY CROWN B) ITS SO EASY!
Luda [366]

Answer:

300

Explanation:

3 0
2 years ago
Read 2 more answers
Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard
podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

t_1 = 6 hours

t_2 = 4 hours

Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

5 0
3 years ago
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