This problem can be solved based on the rule of energy conservation, as the energy of the photon covers both the energy needed to overcome the binding energy as well as the energy of ejection.
The rule can be written as follows:
energy of photon = binding energy + kinetic energy of ejectection
(hc) / lambda = E + 0.5 x m x v^2 where:
h is plank's constant = 6.63 x 10^-34 m^2 kg / s
c is the speed of light = 3 x 10^8 m/sec
lambda is the wavelength = 310 nm
E is the required binding energy
m is the mass of photon = 9.11 x 10^-31 kg
v is the velocity = 3.45 x 10^5 m/s
So, as you can see, all the parameters in the equation are given except for E. Substitute to get the required E as follows:
(6.63x10^-34x3x10^8)/(310x10^-9) = E + 0.5(9.11 x 10^-31)(3.45x10^5)^2
E = 6.41 x 10^-16 joule
To get the E in ev, just divide the value in joules by 1.6 x 10^-19
E = 4.009 ev
Answer:
Explanation:
Let m be the mass of a little car and m' be the mass of another car.
We know that,
Force = mass × acceleration
ATQ,
m × a = 2m × a'
a = 2 × a'
So, the acceleration of another little car is equal to 
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You hold a piece of wood in one hand and a piece of iron in the other. Both pieces have the same volume, and you hold them fully under water at the same depth. At the moment you let go of them, which one experiences the greater buoyancy force?<span>
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Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
Answer:
Explanation:
From the data it appears that A is the middle point between two charges.
First of all we shall calculate the field at point A .
Field due to charge -Q ( 6e⁻ ) at A
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
Its direction will be towards Q⁻
Same field will be produced by Q⁺ charge . The direction will be away
from Q⁺ towards Q⁻ .
We shall add the field to get the resultant field .
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
Force on electron put at A
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
