A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55
N and is parallel to the ramp.What is the speed of the block after it travels a distance d = 0.71 m if the block starts from rest? Use work and energy.
1 answer:
Answer:
The answer is 2,416 m/s. Let's jump in.
Explanation:
We do work with the amount of energy we can transfer to objects. According to energy theory:
W = ΔE
Also as we know W = F.x
We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>
Under the force block gains;
W = F.x →
In the second position block has both kinetic and potential energy. Following the law of conservation of energy;
W = ΔE = Kinetic energy + Potantial Energy
W = ΔE =
Here we can find h in the triangle i draw in the picture using sine theorem;
In a triangle
In our situation
→
Therefore
→
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Answer:
We can retain the original diffraction pattern if we change the slit width to d) 2d.
Explanation:
The diffraction pattern of a single slit has a bright central maximum and dimmer maxima on either side. We will retain the original diffraction pattern on a screen if the relative spacing of the minimum or maximum of intensity remains the same when changing the wavelength and the slit width simultaneously.
Using the following parameters: <em>y</em> for the distance from the center of the bright maximum to a place of minimum intensity, <em>m</em> for the order of the minimum, <em>λ </em>for the wavelength, <em>D </em>for the distance from the slit to the screen where we see the pattern and <em>d </em>for the slit width. The distance from the center to a minimum of intensity can be calculated with:
From the above expression we see that if we replace the blue light of wavelength λ by red light of wavelength 2λ in order to retain the original diffraction pattern we need to change the slit width to 2d:
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