Answer:
7.1934 x 10^12 V/m.s
Explanation:
In order to do this exercise, you need to use the correct formula. Besides that, we need to identify our data.
First we have the radius of the plates which are circular, and it's 0.1 m. The current of the loop (I) is 2.0 A, and the radius of the loop is 0.2 m.
Now with this data, we use the next formula:
I = dE/dt Eo A
Where:
dE/dt = rate of electric field
Eo = constant of permittivity of free space
A = Area of circle
Solving for dE/dT:
dE/dt = I / Eo*A
Now, the area of the circle is A = πr²
A = 3.1416 * (0.1)² = 0.031416 m²
Now solving the electric field:
dE/dt = 2 / (8.85x10^-12 * 0.031416)
dE/dt = 7.1934 x 10^12 V/m.s
The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.
Answer:
v_f = 3 m/s
Explanation:
From work energy theorem;
W = K_f - K_i
Where;
K_f is final kinetic energy
K_i is initial kinetic energy
W is work done
K_f = ½mv_f²
K_i = ½mv_i²
Where v_f and v_i are final and initial velocities respectively
Thus;
W = ½mv_f² - ½mv_i²
We are given;
W = 150 J
m = 60 kg
v_i = 2 m/s
Thus;
150 = ½×60(v_f² - 2²)
150 = 30(v_f² - 4)
(v_f² - 4) = 150/30
(v_f² - 4) = 5
v_f² = 5 + 4
v_f² = 9
v_f = √9
v_f = 3 m/s
Answer:
Sound quality
Explanation:
Loudness refers to the property of sound that depends on the amplitude of the sound wave while pitch is the property of sound that depends on the frequency of the sound wave.
Now, when sounds have the same pitch and loudness, it means they possess the same natural frequency and also, they will have similar resonance and standing waves. Despite all these similarities, the quality of these sounds will tend to vary from one to another.
Answer:
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