We are given a series circuit with two light bulbs. In this case, the light bulbs act as resistors in series and the total resistance is:
That is the sum of all the resistances in series in the circuit. To determine the voltage we can use Ohm's law:
Where "R" is the total resistance and "I" is the current in the circuit. Replacing we get:
Answer:
The image distance is 30 cm
image height = - 5 cm
Explanation:
The formula for calculating the image distance is expressed as
1/f = 1/u + 1/v
where
f is the focal length
u is the object distance
v is the image distance
From the information given,
u = 30
f = 15
By substituting these values into the formula,
1/15 = 1/30 + 1/v
1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30
Taking the reciprocal of both sides,
v = 30
The image distance is 30 cm
magnification = image height/object height = - v/u
Given that object height = 5 cm, then
image height/5 = - 30/30 = - 1
image height = - 5 * 1
image height = - 5 cm
that statement is true
a Third class lever applied when the effort place between the load and the fulcrum.
For example, in a forearm serve
Fulcrum : The elbow
Effort : The effort that putted by the biceps muscle
Load : The arm
3, protons are positive and there are 3 positive atoms visible
Answer:
Explanation:
We know that Impulse = force x time
impulse = change in momentum
change in momentum = force x time
Force F = .285 t -.46t²
Since force is variable
change in momentum = ∫ F dt where F is force
= ∫ .285ti - .46t²j dt
= .285 t² / 2i - .46 t³ / 3 j
When t = 1.9
change in momentum = .285 x 1.9² /2 i - .46 x 1.9³ / 3 j
= .514i - 1.05 j
final momentum
= - 3.1 i + 3.9j +.514i - 1.05j
= - 2.586 i + 2.85j
x component = - 2.586
y component = 2.85
