For an object to be seen, light must leave _______ and enter _______.
1 answer:
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Answer:
a). 100.2 MHz (typical frequency for FM radio broadcasting)
The wavelength of a frequency of 100.2 Mhz is 2.99m.
b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)
The wavelength of a frequency of 1070 khz is 280.3 m.
c. 835.6 MHz (common frequency used for cell phone communication)
The wavelength of a frequency of 835.6 Mhz is 0.35m.
Explanation:
The wavelength can be determined by the following equation:
(1)
Where c is the speed of light, is the wavelength and
is the frequency.
Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.
<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>
Then, can be isolated from equation 1:
(2)
since the value of c is . It is necessary to express the frequency in units of hertz.
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.
<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>
<em> </em>
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 1070 khz is 280.3 m.
<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>
⇒
But
Finally, equation 2 can be used:
Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.
Answer:
a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.
Explanation:
So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.
part a)
So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:
When solving for theta, we get that:
so now we can substitute the corresponding values:
Which yields:
but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:
θ=270°-60°=210°
part b)
for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:
The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:
part c)
In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:
and now up stream:
Once we got these two velocities we will now need to find the time to take each trip:
time down stream:
and the time up stream:
so the total time will be:
d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:
t=1.333hr
e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)
f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.
so
t= 1hr
