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igor_vitrenko [27]
3 years ago
11

I will be so thankful if u answer correctly!!​

Physics
1 answer:
polet [3.4K]3 years ago
3 0

(C) 200 N

Explanation:

The acceleration due to gravity on earth g_{E} is given by

g_{E} = G\dfrac{M_{E}}{R_{E}^{2}}

where G = universal gravitational constant

\:\:\:\:\:\:\:\:\:\:\:\:M_{R} = mass of the earth

\:\:\:\:\:\:\:\:\:\:\:\:R_{E} = radius of the earth

Planet Krypton has twice the mass of earth and 3 times the radius so its acceleration due to gravity g_{K} is

g_{K} = G\dfrac{M_{K}}{R_{K}^{2}}

\:\:\:\:\:\: = G\dfrac{(2M_{E})}{(3R_{E})^{2}}

\:\:\:\:\;\:= (\dfrac{2}{9})G\dfrac{M_{E}}{R_{E}^{2}}

or

g_{K} = (\dfrac{2}{9})\:g_{E}

If we multiply both sides by Superman's mass, we get his weights on both planets:

mg_{K} = (\dfrac{2}{9})\:(mg_{E})

W_{K} = (\dfrac{2}{9})\:W_{E} = (\dfrac{2}{9})(900\:N)= 200\:N

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A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
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Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
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