Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
Explanation:
parallel capacitances add directly
Series capacitances add by reciprocal of sum of reciprocals.
Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]
Ceq = [ C ] + [C / 2] + [C / 3]
Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]
Ceq = 11C/6
If it helps or doesn't I'm sorry, but if you even played the game Minecraft just remember it.
Gold, silver, coal, and iron come from ores.
