Answer:
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Explanation:
The diffraction of a slit is given by the expressions
a sin θ = m λ
where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.
sin θ = m λ / a
If this equation
a ’= 2 a
we substitute
2 a sin θ'= m λ
sin θ'= (m λ / a) 1/2
sin θ ’= sin θ / 2
We can use trigonometry to find the width
tan θ = y / L
as the angle is small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
y ’/ L = y/L 1/2
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Answer: hello your question is poorly written attached below is the complete question
answer :
TA = 1.6*10^-24 * 60 * 2, TB = 1.6*10^-24 * ( 60 + 30 ) * 2 -- ( option 1 )
Explanation:
a = 2m/s^2
Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц
Tb - Ta = m₂ a
∴ Tb = m₂ a + Ta
= ( 30 * 1.6 * 10^-24 * 2 ) + ( 60 * 1.6 * 10^-24 * 2 )
= ( 30 + 60 ) * 1.6 * 10^-24 * 2 ц
Answer:
gravityis an invisible force that pulls objects toward each other. So, the closer objects are to each other, the stronger their gravitational pull is. Earth's gravity comes from all its mass .
In your question where the ask is to calculate the charge that the small sphere carries which is the mass of it is 441g moving at an acceleration of 13m/s^2 nad having and electric field of 5N/C. So the formula in getting the charge is mutliply the mass and the quotients of Acceleration and the Electric Field so the answer is 1,146.6
Answer:
The overall velocity of the water when it hits the bottom is:
Explanation:
Use the law of conservation of energy.
Call it instant [1] to the moment when the water is just before reaching the falls.
At this moment its height h is 206 meters and its velocity horizontally is m/s.
At the instant [1] the water has gravitational power energy
The water also has kinetic energy Ek.
Then the Total E1 energy is:
In the instant [2] the water is within an instant of touching the ground. At this point it only has kinetic energy, since the height h = 0. However at time [2] the water has maximum final velocity
So:
As the energy is conserved then
Now we solve for .