Answer:
Emissions per second = 0.36
Explanation:
Please find the attached question
Solution
Given
Let X be the rate of background emission.
X = B/t
Where B = 36
And t = 100
X = 36/100 = 0.36
Answer: 87.9 m
Explanation:
We can solve the problem by using the following SUVAT equation:
where
v = 26.1 m/s is the final velocity of the car
u = 23.06 m/s is the initial velocity of the car
a = 0.85 m/s^2 is the acceleration of the car
d is the displacement
By re-arranging the equation and substituting the numbers, we find the displacement:
Answer:
V = 411.43 V
Explanation:
The two forces as a result of each of the 2 charges are;
F1 = kq1•q/r
F2 = kq2.q/r
Where r = r/2 since we are dealing with potential difference at a point midway between the charges.
q1 = 5 nC = 5 × 10^(-9) C
q2 = 3 nC = 3 × 10^(-9) C
k = 9 × 10^(9) N.m²/C²
r = 35 cm = 0.35m
r/2 = 0.35/2
Thus;
F1 = (9 × 10^(9) × 5 × 10^(-9) × q)/(0.35/2)²
F1 = 1469.39q
F2 = (9 × 10^(9) × 3 × 10^(-9) × q)/(0.35/2)²
F2 = 881.63q
Net force acting midway is;
F_net = F1 + F2
F_net = 1469.39q + 881.63q
F_net = 2351.02q
Now, we know that formula for electric potential is;
V = kq/r
Thus ;
V = Fr/q derived from the earlier equation for force we used.
Where F is F_net.
V = 2351.02q × r/q
V = 2351.02r
Recall that we are dealing with midpoint and r = r/2
Thus;
V = 2351.02 × 0.35/2
V = 411.43 V
If the force equals, for instance, 100 Newtons then 0.866 × 100 = 86.6 Newtons. This is the magnitude of the resultant force vector on the object.