Answer:
v = 17.71 m / s
Explanation:
We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity
v² = v₀² - 2 g (y -y₀)
v² = 0 - 2g (y -y₀)
when it hits the stone the height is zero and part of the height of the seagull I
v² = 2g y₀
v = Ra (2g i)
let's calculate
v =√ (2 9.8 16)
v = 17.71 m / s
Answer:
8.37×10⁻⁴ N/C
Explanation:
Electric Field: This is the ratio of electrostatic force to electric charge. The S.I unit of electric field is N/C.
From the question, the expression for electric field is given as,
E = F/Q.......................... Equation 1
Where E = Electric Field, F = force experienced by the charged balloon, Q = Charge on the balloon.
Given: F = 8.2×10⁻² Newton, Q = 9.8×10 Coulombs = 98 Coulombs
Substitute these values into equation 1
E = 8.2×10⁻² /98
E = 8.37×10⁻⁴ N/C
Hence the Electric Field of the charged balloon = 8.37×10⁻⁴ N/C
Mental and non mental, an ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal.
Answer:
Explanation:
In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.
To find the fundamental mode you use:

n: mode of the sound
vs: sound speed
L: length of the column of air in the tube.
A) The fundamental mode id obtained for n=1:

B) For the 3rd harmonic you have:

C) For the 2nd harmonic:

Answer:
The power will remain the same for a particular load as we are not changing the load. so if we increase the voltage, the current will decrease to make the net power consumed by the load same as before. If we increase the current, the voltage will decrease for making the power same. The power will only change when we changes the load.
Explanation: