1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MatroZZZ [7]
3 years ago
15

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?
Physics
1 answer:
Andrej [43]3 years ago
3 0

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

You might be interested in
What helped provide support for the Plate Tectonic theory in the 1960s?
skad [1K]
D. Mineral and fossil matches from tests done on different continents.
3 0
3 years ago
Read 2 more answers
Two bodies made of the same material have same external dimension and appearance, but one is compact solid and other is hollow.
Bumek [7]

Answer:

Different

Explanation:

The hollow one will expand even more making it have a larger volume then the solid one so they are different

5 0
3 years ago
Which of these statements is true?
9966 [12]
C.) Meiosis involves two cycles of cell division

Hope this helps!
6 0
3 years ago
Read 2 more answers
A burning candle provides :
mixer [17]
This would be B



Hope this helped
8 0
3 years ago
Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su
dybincka [34]
  1. La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
  2. La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua (v), en metros por segundo:

v = \sqrt{\frac{K}{\rho} } (1)

Donde:

  • K - Módulo de compresibilidad, en newtons por metro cuadrado.
  • \rho - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que \rho = 1\times 10^{3}\,\frac{kg}{m^{3}} y K = 2,16\times 10^{9}\,\frac{N}{m^{2}}, entonces la velocidad de las ondas sonoras es:

v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }

v\approx 1469,694\,\frac{m}{s}

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda (\lambda), en metros, mediante la siguiente fórmula:

\lambda = \frac{v}{f} (2)

Donde f es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que v\approx 1469,694\,\frac{m}{s} y f = 1000\,hz, entonces la longitud de onda de las ondas sonoras es:

\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}

\lambda = 1,470\,m

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

6 0
2 years ago
Other questions:
  • radar wave is transmitted and later reflected off an aircraft and recieved 1.4×10^3 sec after being sent out. how far is the air
    7·1 answer
  • Ezra is pulling a sled, filled with snow, by pulling on a rope attached to the sled. The rope makes an angle θ with respect to t
    15·1 answer
  • show answer No Attempt Treated as a projectile, what is the maximum range, in meters, obtainable by a person if he or she has a
    11·2 answers
  • A particle (charge = 40 μC) moves directly toward a second particle (charge = 80 μC) which is held in a fixed position. At an in
    6·1 answer
  • A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen
    12·1 answer
  • Select the correct answer
    5·2 answers
  • I am lonely to death
    11·2 answers
  • Please help ASAP- A book has a mass of 5 kg, a baby has a mass of 5 kg. Which one weighs more?
    11·2 answers
  • A truck brakes from 15 m/s to 2 m/s in 15 seconds. What is its acceleration?
    15·1 answer
  • Burt and ned paddle their canoe 10 miles upstream and 5 miles downstream in 7 hours. if they can paddle 4 mph in calm water, how
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!