An experiment is designed to test what color of light will activate a photoelectric cell the best. The photocell is set in a cir
2 answers:
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.
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Answer:
See explanation
Explanation:
Solution:-
Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in. Hall probes can determine the magnitude of the field. Another version of the right hand rule emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.
Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. Right hand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire and is valid for any current segment.
( See attachments )
- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:
Where,
I : The current,
r : The shortest distance to the wire,
uo : The permeability of free space. = 4π * 10^-7 T. m/A
- Since the wire is very long, the magnitude of the field depends only on distance from the wire r, not on position along the wire. This is one of the simplest cases to calculate the magnetic field strength - B - from a current.
- The magnetic field of a long straight wire has more implications than one might first suspect. Each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for an arbitrary shape current. The Biot-Savart law is written in its complete form as:
Where the integral sums over,
1) The wire length where vector dl = direction of current (in or out of plane)
2) r is the distance between the location of dl and the location at which the magnetic field is being calculated
3) r^ is a unit vector in the direction of r.
Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by
a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00
At x = 1.00 m
= 4J
Kinetic energy = (1/2)mv²
= 18J
Total energy will be =
4J + 18J = 22J
At x = 5
= -0.24J
Kinetic energy =
= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2
= 3.33 m/s
b) magnitude of force when x = 5.0m
At x = 5.0 m
= 0.016N
Answer:
Explanation:
given,
F = 14.1 i + 0 j + 5.1 k
displacement = 6 m
Assuming block is moving in x- direction
we know,
dW = F dx
hence, work done by the force is equal to