Answer:
0.00354 (N)
Explanation:
Convert to metric system:


Formula for gravitational force:

where s is the distance between 2 bodies masses m and M
Substitute the number to the formula above and since the 2 forces are acting in opposite direction, the total net gravitational force on the mass of origin be:






0.119cm/s is the radius of the balloon increasing when the diameter is 20 cm.
<h3>How big is a circle's radius?</h3>
The radius of a circle is the distance a circle's center from any point along its circumference. Usually, "R" or "r" is used to indicate it.
A circle's diameter cuts through the center and extends from edge to edge, in contrast to a circle's radius, which extends from center to edge. Essentially, a circle is divided in half by its diameter.
dv/dt = 150cm³/s
d = 2r = 20cm
r = 10cm
find dr/dt
Given that the volume of a sphere is calculated using
v = 4/3πr³
Consider both sides of a derivative
d/dt(v) = d/dt( 4/3πr³)
dv/dt = 4/3π(3r²)dr/dt = 4πr²dr/dt
Hence,
dr/dt = 1/4πr².dv/dt
dr/dt = 1/4π×(10)²×150
dr/dt = 1/4π×100×150
dr/dt = 0.119cm/s.
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Answer:
d. to allow the force of expanding gases from the gunpowder to act for a longer time
Explanation:
Answer:
The intensity of light from the 1mm from the central maximu is 
Explanation:
From the question we are told that
The wavelength is 
The width of the slit is
The distance from the screen is 
The intensity at the central maximum is 
The distance from the central maximum is 
Let z be the the distance of a point with intensity I from central maximum
Then we can represent this intensity as
Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

if the angle between the the light at z and the central maximum is small
Then 
Which implies that

substituting this into the equation for the intensity
![I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20%5B%5Cfrac%7Bsin%20%5B%5Cfrac%7B%5Cpi%20w%7D%7B%5Clambda%7D%20%5Ccdot%20%5Cfrac%7Bz%7D%7BD%7D%20%20%5D%7D%7B%5Cfrac%7B%5Cpi%20w%20z%7D%7B%5Clambda%20D%5Cfrac%7Bx%7D%7By%7D%20%7D%20%7D%20%5D)
given that 
We have that
![I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2](https://tex.z-dn.net/?f=I%20%3D%20I_o%20%5B%5Cfrac%7Bsin%5B%5Cfrac%7B3.142%20%2A%200.45%2A10%5E%7B-3%7D%7D%7B%28620%20%2A10%5E%7B-9%7D%29%7D%20%5Ccdot%20%5Cfrac%7B1%2A10%5E%7B-3%7D%7D%7B3%7D%20%5D%7D%7B%5Cfrac%7B3.142%20%2A%200.45%2A10%5E%7B-3%7D%2A1%2A10%5E%7B-3%7D%20%7D%7B620%2A10%5E%7B-9%7D%20%2A3%7D%20%7D%20%5D%5E2)
![=I_o [\frac{sin(0.760)}{0.760}] ^2](https://tex.z-dn.net/?f=%3DI_o%20%5B%5Cfrac%7Bsin%280.760%29%7D%7B0.760%7D%5D%20%5E2)
