Answer:

Explanation:
t = Time taken = 
i = Current = 3 A
q(0) = Initial charge =
Charge is given by

The magnitude of the net electric charge of the capacitor is 
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
q = f₁
2) for an object located at p = 25 cm
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
= 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = 
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
Answer:
<em>Well, I think the best answer will be is </em><em>1.59 g/mL Good Luck!</em>
Answer:
Vi = 5 m/s
Explanation:
let (a) acceleration = 0.75 m/s²
(t) time = 20 seconds
Vf = final velocity = 72 km/hr (convert to m/s to units consistency = 20 m/s)
find Initial velocity (Vi)
Vf - Vi
a = -----------
t
Vi = Vf - (a * t) = 20 - (0.75 * 20)
Vi = 5 m/s
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU