Light of wavelength 660 nm passes through two narrow slits 0.65 mm apart. The screen is 2.55 m away. A second source of unknown
1 answer:
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a) The work done is 920 J
b) The work done is 5200 J
Explanation:
a)
In this first part of the problem, the crate is moved horizontally at constant speed.
The work required in this case is given by
where
F is the magnitude of the force applied
d is the displacement of the crate
is the angle between the direction of the force and of the displacement
Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so
F = 230 N
The displacement is
d = 4.0 m
And the angle is , since the force is applied horizontally. Therefore, the work done is
b)
In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore
F = W = 1300 N
The displacement this time is again
d = 4.0 m
And the angle is , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is
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Answer:
Explanation:
We need 2 different equations for this problem: first the velocity of sound equation, then the frequency of the sound equation.
The velocity of sound is found in:
v = 331.5 + .606T
We need to find that first in order to fill it into the frequency equation which is
where v is the velocity we will find the part a, f is frequency and lambda is the wavelength. Starting with the velocity of the sound:
v = 331.5 + .606(25) and
v = 331.5 + 15 and rounding correctly using the rules for sig fig when adding:
v = 347 m/s
Filling that into the frequency equation:
and
so