Answer: 4.27 *10^6 N/C
Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:
E=k*x/(a^2+x^2)^3/2 where a is the ring radius and x the distance to the point measured from the center of the ring.
Replacing the data we have:
E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2
then
E=4.27 * 10^6 N/C
Answer:
2274 J/kg ∙ K
Explanation:
The complete statement of the question is :
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
= mass of metal = 400 g
= specific heat of metal = ?
= initial temperature of metal = 100 °C
= mass of aluminum cup = 100 g
= specific heat of aluminum cup = 900.0 J/kg ∙ K
= initial temperature of aluminum cup = 15 °C
= mass of water = 500 g
= specific heat of water = 4186 J/kg ∙ K
= initial temperature of water = 15 °C
= Final equilibrium temperature = 40 °C
Using conservation of energy
heat lost by metal = heat gained by aluminum cup + heat gained by water

That depends on what type of pressure you are attempting to measure, to measure Atmospheric pressure, you would use a Barometer. To measure things like tires, you could use a Tire Pressure Gauge. For Industrial processes and boilers, you would use a Manometer. For pressure vessels, you would use a Bordon Gauge. <span />
Hello!
This is a matter of superposition.
When the waves peak at the same time and place, they produce constructive interference, meaning the waves interact together in a positive way, to make a wave with Amplitude of both waves added together. When the peaks differ however, at the same time and place, then it is destructive interference and the waves essentially cancel each other out.
Hope this helps. Any questions please just ask. Thank you kindly.