Two stars in the night sky make an angle of 10^-3 radians at the eye of Matilda. If Arthur, who can just distinguish two spots, 0.5 cm apart, at a distance of 10 m views the stars, can he distinguish them as two distinct objects?
Answers
Let's calculate the angle between two spots, 0.5 cm apart,
that are 10m away from you.
1 m = 100 cm
10m = 1,000 cm
The angle between the two spots is very very close to the angle
whose tangent is
0.5cm/1,000cm = 0.0005 .
That angle is tan⁻¹(0.0005) = 0.0005 radian = 0.029 degree .
Matilda is looking at two stars that appear to be 0.001 radian apart.
That's (0.001 / 0.0005) = 2 = TWICE as far apart as Arthur can distinguish.
As long as Arthur and Matilda are on the same planet while looking at
these two stars ... or as long as they're even in the same solar system ...
Arthur will have no trouble separating them.
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What kind of scientist investigation involve making observations
B. potential energy
C. volume
D. mass
Answers
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the first order bright band
A= 0.78
B= 0.11
C= 0.032
D= 0.62
Answers
The angle between the central bright band and the first order bright band is 0.78⁰, hence option A is correct.
According to the question:
A light with a wavelength of 410 nm falls on two slits separated by 0.03 mm,
So, the angle between the central bright band and the first order bright band is,
The angle will be, θ = λ/d radian
λ = 410 nm (410 × 10 ⁻⁹ m)
d = separation of slits (0.03 mm) or 0.03 × 10 ⁻³ m
θ =
= 0.01367 radian
= 0.01367 × (180/π)⁰
= 0.78⁰
Thus, the angle between the central bright band and the first order bright band is 0.78⁰, hence option A is correct.
Learn more about light, here:
#SPJ2
Answer:
B
Explanation:
2.4 m/s
0.42 m/s
1.9 m/s
Answers
A water wave travels 36 meters in 15 seconds. What is the speed of the wave?
2.4 m/s
Answers
For this case, the first thing we are going to do is define a reference system.
We define the vertical axis of the reference system, pointing upwards.
Therefore, doing summation of forces in vertical direction we have:
Then, since we want the net force to be equal to zero, then:
From here, we clear the force that we must apply on the object:
The positive sign means that the force points upwards.
Answer:
to create a net force of zero on the object you must apply:
33 Newton upwards.
Answers
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Answers
(output work) = (output force) x (distance)
450 N-m = (output force) x (3 meters)
Divide each side
by 3 meters: Output force = (450 N-m) / (3 m)
= 150 newtons .
With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.
It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m). So everything
is nice and hunky-dory.
___________________________________
Well, my goodness !
I didn't even need to go through all of that.
Given:
-- The input force to the machine is 50 newtons.
-- The mechanical advantage of the machine is 3 .
That right there tells us that
-- The output force of the machine is 150 newtons.
We don't need any of the other given information.