If you shine a single light source on a screen you will see that the entire screen is lit up. Assume this light source is of a s
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Answer:
The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters
Explanation:
Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;
v² = u² - 2·g·h
Where;
v = The final velocity
u = The initial velocity
g = The acceleration due to gravity ≈ 9.8 m/s²
h = The height she jumps
At the maximum height, = 0.500 m, she jumps, v = 0, therefore, we have;
0² = u² - 2·g·
u² = 2 × 9.8 × 0.5 = 9.8
u = √9.8 ≈ 3.13
u = 3.13 m/s
Her initial jumping velocity ≈ 3.13 m/s
Escape velocity,
Where;
M = The mass of the asteroid
G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)
r = The radius of the asteroid
The average density of the Earth = 5515 kg/m³
The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³
The escape velocity, she has, ≈ 3.13 m/s is therefore;
Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.
In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
where t is the time,
is the maximum charge on the capacitor at voltage V, and 
is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.
1) Using
and 
, the time constant of the circuit is:
2) To find the charge on the capacitor at time
, we must find before the maximum charge on the capacitor, which is
And then, the charge at time is equal to
3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be
. We can see this also from the previous formule, by using 
:
where t is the time,
Using this law, we can answer all the three questions of the problem.
1) Using
2) To find the charge on the capacitor at time
And then, the charge at time
3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be