A photoelectric cell is an electronic device which is used to convert light energy into electric energy.The operation of this device is based on photoelectric effect.
Light of suitable frequency i.e greater or equal to threshold frequency will fall on the cathode maintained at negative potential.The electron emission will take place and these electrons are drifted towards the anode which is at positive potential.
Here,only those radiations will be capable of emitting electrons irrespective of surface barrier of metals whose energy is greater than the work function.
We know that the radiation having long wavelength has least energy as energy and wavelength are inversely proportional to each other.

Here h is the Planck's constant,c is the velocity of light.
Here we have been given red light and blue light.
In the visible spectrum of radiation, the red light has longer wavelength than all other colors of light.Hence blue light has more energy as it's wavelength is less as compared to red light.
Hence, the blue light will activate the most and red the least.
Answer:
negative at first.
Explanation:
It's important to disconnect the negative side at first, otherwise you can cause an electrical short if positive is removed first.
the upward force is from the table and the downward is from gravity they are equal in force so the box doesn't fly up or sink down
Answer:
(a) 
(b) 
Explanation:
<u>Electric Circuits</u>
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:

(a) The electromagnetic force of the battery is
and its internal resistance is
. Knowing the equivalent resistance of the headlights is
, we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:

Solving for i

i=2.28\ A
The potential difference across the headlight bulbs is


(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is
