Answer: Here is the complete question:
A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 103 N/C. If the ball is in equilibrium when the string makes a 30 angle with the vertical, what is the net charge on the ball?
Answer: The charge on the ball is 5.71 × 10^-4 C
Explanation:
Please see the attachments below
Answer: ![-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BBr%5E.%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Explanation:
Rate of a reaction is defined as the rate of change of concentration per unit time.
Thus for reaction:

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{d[Br^.]}{2dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7Bd%5BBr%5E.%5D%7D%7B2dt%7D)
or ![Rate=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Thus ![-\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BBr%5E.%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7Bd%5BBr_2%5D%7D%7Bdt%7D)
Answer:
The surface gravity g of the planet is 1/4 of the surface gravity on earth.
Explanation:
Surface gravity is given by the following formula:

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:


The problem tells us the radius of the planet is twice that of the radius on earth, so:

If we substituted that into the gravity of the planet equation we would end up with the following formula:

Which yields:

So we can now compare the two gravities:

When simplifying the ratio we end up with:

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.
Solution :
Speed of the air craft,
= 262 m/s
Fuel burns at the rate of,
= 3.92 kg/s
Rate at which the engine takes in air,
= 85.9 kg/s
Speed of the exhaust gas that are ejected relative to the aircraft,
=921 m/s
Therefore, the upward thrust of the jet engine is given by

F = 85.9(921 - 262) + (3.92 x 921)
= 4862635.79 + 3610.32
= 
Therefore thrust of the jet engine is
.
Assuming that reaching a height 0 doesn’t stop the ball, and that it accelerates at 9.8 m/s^2, the ball would be traveling at 0.5 + 0.7*9.8 = 7.36 m/s downwards.