T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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F = m • a
What we know:
- Gravity: 9.8 m/s
- Force: 490 N
Equation derived:
m = F/a
m = 490/9.8
= 50 kg
Explanation:
90 kmhr—1 x 1000/3600 = 25ms—1
U = 0 ms—1
V = 25ms—1
t = 10 s
a = ?
a = V - U/t
a = 25 - 0/10
a = 25/10
a = 2.5 ms—1
Answer:
0.69
Explanation:
Sum of the forces on the sled in the y direction:
∑F = ma
Fn - 52 N = 0
Fn = 52 N
Sum of the forces on the sled in the x direction:
∑F = ma
36 N - Fn μ = 0
Fn μ = 36 N
μ = 36 N / Fn
Substitute:
μ = 36 N / 52 N
μ = 0.69
Answer:10 km westExplanation:he go 40 east then 50 west 50-40 is 10 so he displaces 10 km and as west is more than east in terms of km so we will say that it's 10 km west pls mark as brainliest thanks
