Atmospheric electricity and storms, electrostatic control filters, and industrial electrostatic seperation as well as spark discharge. these are just a few. hope it helps.
Answer:
300 cos 30 = 40 a + 40 * .2 * 10
Total force = mass * acceleration + frictional force
260 = 40 a + 80
a = 180 / 40 = 4.5 m/s^2
Check:
15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)
T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block
260 - 97.5 = 162.5 net force on 25 kg block
162.5 = 4.5 (25) + 25 * 10 * .2
162.5 = 112.5 + 50 = 162.5
4.5 m/s^2 checks out as correct
Answer:
(a) 21.36 ohms
(b) 5.62 A
Explanation:
Parameters given:
Potential difference, V = 120 V
Power, P = 674 W
(a) Power is given as:
P = V²/R
Where R is resistance
=> R = V²/P
R = 120²/674
R = 14400/674
R = 21.36 ohms
(b) Power is also given as:
P = I*V
Where I = Current (time rate of flow of Electric charge)
=> I = P/V
I = 674/120
I = 5.62 A
Answer:
La longitud del camino recorrido es de 25.9 [m]
Explanation:
Se reemplaza el valor de tiempo en segundos en la ecuación dada de desplazamiento
x=10+20*(3) - 4.9*(3)^2
x= 25.9 [metros]
Answer:
(A) ratio of electric force to weight will be 
(b) Electric field will be 
Explanation:
We have given mass of bee = 100 mg = 
Charge on bee 
Electric field E = 100 N/C
Weight of the bee 
Electric force on the bee 
So the ratio of electric force on the bee and weight is 
(B) To hold the bee in air electric force must be equal to weight of bee
So 
