Answer:
Q1 = 7.25*10^(-16) C
Explanation:
We are given;
electric field strength = (1 x 10^5 N/C
drag force (F) = 7.25 x 10^(-11) N
The question says it's moving with constant velocity. This means that he particle is in equilibrium and not accelerating.
Columbs law force of attraction or repulsion between two charges is given as;
F=(KQ1Q2)/r²
Now, electric field strength is given as the formula;(K*Q2)/r², thus plugging the relevant values gives us;
7.25 x 10^(-11) N= (1 x 10^(5) N/C)Q1 Q1 = 7.25 x 10^(-11) /(1 x 10^(5))
Q1 = 7.25*10^(-16) C
A) average acceleration = final velocity - initial velocity / time
= 7700 - 0 / 11
= 700ms^-2
B) force = mass x acceleration
= (3.05 x 105) x 700
= 320.25 x 700
= 224,175N
Explanation:
PRIMERO HACES EL RECUENTO DEL TIEMPO Y LO CONVIERTES EN
SEGUNDOS Y ENTONCES
<em>t</em> = 227 s 
= 227 S - 38 s = 189 s
= 38 s
LUEGO USANDO LA ECUACIÓN DE GALILEO GALILEI SSUPONIENDO
QUE EL MOVIL VIAJA A VELOCIDAD CONSTANTE
<em>v</em> = 3.50 m/189 s = 0.0185 m/s
PARA LA DISTANCIA NTRE B Y C
= 0.0185 m/S( 38 s) = 0.703 m
LA HORA EN QUE EL MOVIL PASA POR A ES
11:43:15 - 38 s - 189 s = 11:39:29
about 5 watts (5W) of power
Answer:
4.0 m/s
Explanation:
acceleration = (16-2) /3.5 = 4 m/s
