A 110 V power line is protected by a 15 A fuse. What is the maximum number of 400 W lamps that can be simultaneously operated in
1 answer:
Answer:
Total number of lamps will be 4
Explanation:
We have given power of the lamp W = 400 watt
Potential difference across the lamp V=110 volt
We know that power is equal to
So
Total current is given 15 A
As it is given that lamps are connected in parallel so total current is the sum of current through each lamp
So number of lamp will be
As the lamp can not be in negative
So total number of lamps will be 4
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Solution:
In this question we have given,
initial velocity,u = 0
final velocity, v = 1m/s
displacement =.5mm (Because 1mm=.001m)
We have to find
1.acceleration
2. Time
1. we know third equation of motion is given as,
............(1)
Put values of v,u and s in equation (1)
2. We know First equation of motion is given as
......................(2)
put values of v, u and a in equation(2)
Answer:
Focal length of the eye-lens is 2.0 cm
Explanation:
Given:
An eye-lens system which will follow converging lens (convex lens) sign conventions and knowing that lens-retina distance is the image distance.
Object distance, =
Image distance, =
cm
We have to find the focal length.
Lets say that the focal length is .
Lens Formula:
⇒
Sign-convention:
We will follow the Cartesian coordinate,placing the lens at the origin of the plane, optical center is at now as the object will be towards negative x-axis so it is negative others are on the right side of the origin so they are positive.
Using the lens formula:
⇒
⇒
⇒ <em>...as </em>
a<em>nd there is no </em>
<em> so it is zero only.</em>
⇒ cm
So the focal length is 2.0 cm with two significant figures.