There may be an improper voltage in the device.
<u>Explanation</u>:
- Given the device has unknown circuitry and the voltage across the device crosses the voltage across the resistor at 500 Hz. In this device, the current flow is not correct. It is improper.
- The elements in the device are ammeter, capacitor, inductor, voltmeter, and battery. The components in the device may not be connected properly. This is due to the improper flow of current and condition of the voltmeter.
- So these are the type of elements in the device.
Answer:
It will be cut in half
Explanation:
The diffraction of a slit is given by the formula
a sin θ = m where
a = width of the slit,
λ = wavelength and
m = integer that determines the order of diffraction.
Next we divide both sides by a, we have
sin θ = m λ / a
Also, recall that
a’ = 2 a
Then we substitute in the previous equation
2asin θ' = m λ, if divide by 2a, we have
sin θ' = (m λ / 2a).
Now again, from the first equation, we said that sin θ = m λ / a, so we substitute
sin θ ’= sin θ / 2
Then we use trigonometry to find the width, we say
tan θ = y / L
Since the angle is small, we then have
tan θ = sin θ / cos θ
tan θ = sin θ, this then means that
sin θ = y / L
we will then substitute
y’ / L = y/L 1/2
y' = y / 2
this means that when the slit width is doubled the pattern width will then be halved
The gravitational force between the Earth and the satellite (its "weight") is inversely proportional to the distance between the centers of both objects.
On the surface, their centers are separated by 1 Earth radius.
12,000 miles above the surface, they're separated by 4 Earth radiii.
(4/1) = 4
So after the move, the satellite's weight is (1/4²) = 1/16 of its surface weight.
(321 lb) / (16) = (20 and a hair) lb
The correct choice from the given list is " <em>>20 lb "</em> .
Since you are referring to the TI-203 and TI-205, you need to know the actual masses of these two isotopes. TI-203 has 202.9723 amu and TI-205 has 204.9744 amu. Since you are concluding that this Thallium have 29.5% (Ti-203) and 70.5% (Ti-205), you need to multiply the percentage to the actual masses of the isotopes. With that, you should be able to get 204.3833 amu
