Answer:
1/9
Explanation:
<em>Newton’s Law of Universal Gravitation
</em>
Objects with mass feel an attractive force that is proportional to their masses and inversely proportional to the square of the distance.
F = GMm/r²
where
F
- the gravitatioal force in Newtons,
M and m -two masses in kilograms
r - the separation in meters.
G - the gravitational constant (6.674*10
⁻¹¹ N
(m/kg)
²
)
Because of the magnitude of G , gravitational force is very small unless large masses are involved.
So according to above equation , when the masses are not changing , force is inversely propotional to the square of distance
F1 ∝ 1/r² ---------------(1)
F2 ∝ 1(3r)²
F2 ∝ 1/9r²--------------(2)
(2)/(1)
From their you get as the distance tripled, Force reduce by a factor of 9(3³)
for example , assume the distance get doubled ,Force reduce by a factor of 4 (2²)
Answer:
D. High frequency and short wavelengths.
Explanation:
If a wave is high in energy it will have a higher frequency.
High frequency = short wavelengths
(a) The ball's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :
0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )
<em>t</em> = 0 or (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 <em>g t</em>
<em>t</em> = (40 m/s) sin(40º) / <em>g</em>
<em>t</em> ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So
0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>
where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :
<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)
<em>y</em> ≈ 8.4 m
