Answer:
The answer to your question is m₂ = 38.5 kg
Explanation:
Data
distance = d = 2.1 x 10⁻¹ m
Force = 3.2 x 10⁻⁶ N
m₁ = 55 kg
m₂ = ?
G = 6.67 x 10 ⁻¹¹ Nm²/kg²
Process
1.- To solve this problem use Newton's law of Universal Gravitation.
F = G m₁m₂ / r²
-Solve for m₂
m₂ = Fr² / Gm₁
2.- Substitution
m₂ = (3.2 x 10⁻⁶)(2.1 x 10⁻¹)² / (6.67 x 10⁻¹¹)(55)
3.- Simplification
m₂ = 1.411 x 10⁻⁷ / 3.669 x 10⁻⁹
4.- Result
m₂ = 38.5 kg
Ball thrown into the air at an angle.
Answer:
Approximately 
(assuming that the projectile was launched at angle of 
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing 
is the same as the altitude 
at which this projectile was launched: 
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 
(upwards,) the vertical velocity right before landing would be 
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be 
. In other words, 
.
Hence, the time it takes to achieve a (vertical) velocity change of 
would be:
.
Hence, this projectile would be in the air for approximately .
Answer:
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Answer:
Though the question is not specified here, but this information can determine the following quantity: period T= 6 secs, Frequency F=1/6 Hz, speed of rotation V= 2 pi ft/sec and wave length =pi/3 ft
Explanation:
