In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
where D=5.00 m is the distance of the screen from the slits, and
is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
And from the relationship between frequency and wavelength,
, we can find the frequency of the light:
When I see the word "which" at the beginning of your question,
I just KNOW that there's a list of choices printed right there
next to he part that you copied, and for some mysterious
reason, you decided not to let us see the choices.
Any flashlight, light bulb, laser, or spark ... like lightning ...
converts some electrical energy into some light energy.
<span>The velocity would be 54.2 m/s
We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
Answer:
2.47 m
Explanation:
Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.
The horizontal velocity of the ball is constant:
and the time taken to cover the horizontal distance d is
So this is the time the ball takes to reach the horizontal position of the crossbar.
The vertical position of the ball at time t is given by
where
is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:
The height of the crossbar is h = 3.05 m, so the ball passes
above the crossbar.
