At 8:00 pm, the velocity of the storm is 55 mi northeast. Assuming that the direction is exactly northeast, the angle is 45°
At 11:00 pm, the velocity is 75 mi north. The angle is 90°
In vector form
55 ∠ 45°
and
75 ∠ 90°
The magnitude and direction of the average velocity is
(55 ∠ 45° + 75 ∠ 90° ) / 3
Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.
<h3>How to find Wavelength of the light?</h3>
When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.
This is a straightforward situation in which we can apply the
Fraunhofer single slit diffraction equation:
y = mλD/a
Where:
y = Displacement from the center line for minimum intensity = 1.35 mm
λ = wavelength of the light.
D = distance
a = width of the slit = 0.75
m = order number = 1
Solving for λ
λ = y + a/ mD
Changing the information that the issue has provided:
λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2
=5.0625 *10^-7 = 506.25
so
Wavelength of the light 506.25.
To learn more about Wavelength of the light refer to:
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Answer:
Yes! Thinking about it graphically a position vs time graph models meters per second in most cases, making every point on the line have the units m/s. If we want the find the slope we are finding the change between each point and those units would change to m/s/s or m/s^2 giving us the same units for acceleration. Simply put, slope of a velocity graph gives us acceleration.
Explanation:
1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].
2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].
3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].
4. Trick question? If the circuit becomes open, then no current will flow. Moreover, even if the voltmeter were kept as element of the circuit, voltmeters generally have a very high resistance (an ideal voltmeter has infinite resistance), so the current moving through the circuit will be negligible if not nil. In any case, the ammeter reading would be 0 A [choice B].
C.half the energy is carried by the electric field and half is carried by the magnetic field.
