i believe the answer is 1.5e+6
hope this helps!
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
= 
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
Explanation:
The two postulates of special theory of relativity
Postulate 1: The law of physics are invariant under any of inertial frame of reference.
Postulate 2: The velocity of light is remains same in each ans every frame of reference and independent of relativity.
They are differ from classical mechanics that in classical mechanics there is no change in mass and length in relative velocity but in relativistic mechanics it changes.
These two postulates implements in phenomenon like time dilation , length contraction etc.
Thanks
Given parameters:
Mass of the body = 200g
Force on the body = 10N
Unknown parameters:
Acceleration produced by the force = ?
To solve this problem we must first define force in terms of mass and acceleration. This is possible due to the Newton's first law of motion.
Force = mass x acceleration
Here the unknown is acceleration and we can easily solve for it.
But we must take the mass to kilogram in order for it to cancel out.
1000g = 1 kg
200g = x kg =
= 0.2kg
Now input the parameters and solve;
10 = 0.2 x acceleration
Acceleration =
= 50m/s²
The acceleration produced by the body is 50m/s²
Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ

Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.