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]
2 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]2 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
10 PTS!
spin [16.1K]
" This manual applies to Compact Liquid <span>Fuel Pumps & </span>Dispensers<span> The </span>liquid pressure<span> range is from 0.5 - 20m These totals </span>can<span> be displayed by </span>pressing<span> the CLEAR </span>button<span> on the preset keypad five times in When connecting to sites </span>powered<span> by. "</span>
5 0
3 years ago
Read 2 more answers
How much larger is the diameter of Jupiter compared to the diameter of Saturn?
allsm [11]
Really long we’ll not long but far in distance
5 0
2 years ago
An object has a mass of 10.2 kg, what is its weight in Newton's?
Juliette [100K]

Answer:

The answer is 98 N

Explanation:

I am pretty sure

3 0
2 years ago
Read 2 more answers
What is the acceleration of a 349 kg object that moved with a force of 750 N?
zavuch27 [327]

Answer:

<h3>The answer is 2.15 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

f is the force

m is the mass

From the question we have

a =  \frac{750}{349}  \\  = 2.14899713...

We have the final answer as

<h3>2.15 m/s²</h3>

Hope this helps you

4 0
2 years ago
A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te
Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

\Delta L=\alpha L_{0}\Delta T

Where:

  • L(0) is the initial length
  • ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
  • ΔL is the final length minus the initial length (L(f)-L(0))
  • α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)  

So, we have:

L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})

L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})

L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)

L_{f}=0.117\: m

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0
3 years ago
Other questions:
  • Which of the following statements is true? The atmosphere is an example of a solution. In a glass of sugar water, sugar is the s
    5·2 answers
  • What is the electric power used by an appliance if the current is 2 A and the voltage is 50 volts? Show work and include all uni
    11·1 answer
  • Ali pushes a toy car initially at rest towards javed by exerting a constant horizontal force F of magnitude 10N through a distan
    10·1 answer
  • Why do solids have a definite shape?
    13·2 answers
  • A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
    12·1 answer
  • It is observed that the number of asteroids (or meteoroids) of a given diameter is roughly inversely proportional to the square
    13·1 answer
  • What happens to the force between charged and uncharged objects as the distance between them decreases?
    15·1 answer
  • How many minutes will it take a car to go from a stop to 33 km/hr if it accelerates at 10 km/hr²
    6·2 answers
  • 12. Since the Clean Air Act was passed in 1970, why is the air still not clean?
    10·1 answer
  • How far apart are two conducting plates that have an electric field strength of 4. 4 kv/m between them, if their potential diffe
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!