Answer:
x component 3.88 y- component 14.488
Explanation:
We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure
Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )
y- component will be 15 sin 75°=14.488
For verification the resultant of x and y component should be equal to 15
So
Explanation:
Formula depicting relation between total flux and total charge Q is as follows.
(Gauss's Law)
Putting the given values into the above formula as follows.
Q =
=
=
= -8.4 nC
Therefore, when the unknown charge is q then,
-14.0 nC + 33.0 nC + q = -8.4 nC
q = -27.4 nC
Thus, we can conclude that charge on the third object is -27.4 nC.
Answer:
The doppler effect equation is:
In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:
v = f*λ
then:
f = v/λ
and v is the speed of light, then:
f = c/λ
where:
f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm
f is the real frequency, in this case, is (3*10^17nm/s)/650 nm
vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.
v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s
v0 is your speed, this is what we want to find.
Replacing those quantities in the equation, we get:
(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm
(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)
1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)
1.182*(3*10^8 m/s) - (3*10^8 m/s) = v0 = 54,600,000 m/s
So your speed was 54,600,000 m/s, which is a lot.
Answer:
.
Explanation:
The frequency of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)
The wave in this question travels at a speed of . In other words, the wave would have traveled
in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be
How many wave cycles can fit into that ? The wavelength of this wave
gives the length of one wave cycle. Therefore:
.
That is: there are wave cycles in
of this wave.
On the other hand, Because that of this wave goes through that point in each second, that
wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be
Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:
.
The calculations above can be expressed with the formula:
,
where