Answer:
The force becomes 16 times what it is now.
Explanation:
The formula for gravitational force is
F = G * m1 * m2 / r^2
When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.
If you double m1 and m2 and don't do anything to r, you've already increased the force by 4 times. (2m1 * 2m2 = 4 * m1 * m2)
But you are not finished. If you 1/2 the distance, you are again increasing the Force by 4 times. 1 / (2r) ^2 = 1/ 4* r^2
Because this is in the denominator, the 1/4 is going to flip to the numerator.
So the total increase is going to be 4 * (4 * m1 * m2) = 16 * m1 * m2.
Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.
Newtons third law of motion states that for every action, there is an equal an opposite reaction. This means that the force on back on something is going to be equal in size and opposite in direction.
Answer:
horizontal component = 10.54m/s
Explanation:
horizontal component = 13.2cos37°
horizontal component = 10.54m/s
Can't
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Answer:
a) t = 0.90 s, b) t = 0.815 s, c) t = 0.90 s, d) x = 3.6 m, e) t = 0.639 s
Explanation:
all these exercises are about kinematics
a) The body is released from rest,
y = y₀ + v₀ t - ½ g t²
in this case when reaching the ground y = 0 and its initial velocity is vo = 0
0 = y₀ + 0 - ½ g t²
t² = 2 y₀ / g
t² = 2 4 /9.81
t² = 0.815
t = √0.815
t = 0.90 s
b) It is thrown upwards at v₀ = 4 m / s
y = y₀ + v₀ t - ½ g t²
in this case the initial and final height is the same
y = y₀ = 0
0 = v₀ t -1/2 g t²
t = 2 v₀ / g
t = 2 4 /9.81
t = 0.815 s
c) the ball is at y₀ = 4 m and its initial velocity is horizontal v₀ = 4 m / s
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ g t²
t² = 2 i / g
t² = 2 4 / 9.81
t² = 0.815
t = 0.90 s
d) the horizontal distance traveled is
x = v₀ₓ t
x = 4 0.90
x = 3.6 m
e) We can calculate the time to fall from I = 2 m
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ g t²
t² = 2 y₀i / g
t² = 2 2 /9.81
t² = 0.4077
t = 0.639 s
Therefore, when making measurements, you should find readings around this value.
