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<span>By pythagorean theorem then, the vertical side of the right triangle must be 12.
Then if x is the angle between the horizontal side and the hypotenuse, sin(x) = 12/13 but also the anser should be in this sentences.
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Answer:
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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
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