Answer:
1m/s is the acceleration used. C
Explanation:
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To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.
Kepler's third law tells us that

Where
T= Period
G= Gravitational constant
M = Mass of the sun
a= The semimajor axis of the comet's orbit
The period in years would be given by

PART A) Replacing the values to find a, we have




Therefore the semimajor axis is 
PART B) If the semi-major axis a and the eccentricity e of an orbit are known, then the periapsis and apoapsis distances can be calculated by



Newton's third law states that for every action, there is an equal and opposite reaction. When you let go of the ballon, you are letting the force out but the force also pushes the balloon back.
Answer:
121550 J
Explanation:
Parameters given:
Mass, m = 0.34kg
Specific heat capacity, c = 14300 J/kgK
Change in temperature, ΔT = 25K
Heat gained/lost by an object is given as:
Q = mcΔT
Since ΔT is positive in this case and also because we're told that heat was transferred to the hydrogen sample, the hydrogen sample gained heat. Therefore, Q:
Q = 0.34 * 14300 * 25
Q = 121550J or 121.55 kJ
a. There is nothing suggesting that the balloon is accelerating horizontally, so we can assume that its horizontal speed is constant. The time before the balloon crash into the hill is simply the distance between the balloon and the hill divided by its velocity. Remember that velocity is simply the amount of distance that a object travels in a certain amount of time:

b. Know that you know the maximum amount of time that the balloon can take to gain 28m of altitude, the minimum acceleration can be found using the equations constant acceleration motion:

where a is the acceleration, v_o is the initial vertical velocity, 0 as the balloon is not moving vertically before starting to ascend. xo is the initial position, which we will give a value of 0m.

c. As we said before, there isn't any kind of force that accelerates the balloon horizontally, therefore, we can consider that its horizontal velocity is constant and equal to 2.8m/s
d. Acceleration is the amount of change in velocity after a given amount of time. So, with the acceleration and the time we can fin the velocity:
