Answer:
2.9 E14 Hz
Explanation:
As we know by Einstein's equation that energy incident on the photo sensitive surface will be used by the surface to eject electron out of the surface with some kinetic energy.
This is given by
now the threshold frequency is the minimum frequency of the incident photons due to which electrons are ejected out with minimum kinetic energy or least kinetic energy.
So here when KE = 0 in the graph then corresponding to that position the frequency will be given as threshold frequency
so here from graph when KE = 0
Answer:
The average net force on the truck is 375 Newtons.
Explanation:
Using Newton's 3rd equation of motion, we have :
×a×s
where, v = final velocity = 25 m/s
u = initial velocity = 20 m/s
a = acceleration
s = distance traveled = 300 m
Using these values in the above equation, we get acceleration = 0.375 m/
Using Newton's second law, we have:
F=m×a
where m = mass = 1000 kg
a= acceleration = 0.375 m/
Putting values we have F=375 N
A front is a narrow region between two air masses of different densities.
When light passes from a faster medium into a slower medium, light will be refracted toward a line drawn perpendicular to the point of refraction. <em>(B)</em>
Answer:
The magnitude of magnetic field at given point = 
× 
T
Explanation:
Given :
Current passing through both wires = 5.0 A
Separation between both wires = 8.0 cm
We have to find magnetic field at a point which is 5 cm from any of wires.
From biot savert law,
We know the magnetic field due to long parallel wires.
⇒ 
Where 
magnetic field due to long wires, 
, 
perpendicular distance from wire to given point
From any one wire 
5 cm, 
3 cm
so we write,
∴ 
![B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]](https://tex.z-dn.net/?f=B%20%3D%5Cfrac%7B%204%5Cpi%20%5Ctimes10%5E%7B-7%7D%20%5Ctimes5%7D%7B2%5Cpi%20%7D%20%5B%5Cfrac%7B1%7D%7B0.03%7D%20%2B%20%5Cfrac%7B1%7D%7B0.05%7D%20%5D)
Therefore, the magnitude of magnetic field at given point = 
