<span>C.CT scan, X-ray imaging, MRI
</span>
Answer:
Explanation:
Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:
Add them all together to get the x component of the resultant vector, V:
Do the same to find the y components of the part of this journey:
Add them together to get the y component of the resultant vector, V:
One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.
We find the final magnitude:
and, rounding to 2 sig dig's as needed:
1.0 × 10² m; now for the direction:
58°
Answer:
Speed = 0.296m/2
Period = 0.203 s
Explanation:
If by 'long' you mean the wavelength of the waves, then the wavelength 
.
The frequency 
of the waves is 14.8 waves every 3 seconds or
.
Now the relationship between wavelength 
, frequency and speed 
of the waves is:
We put in the values and 
and get:
Now the period 
is just the inverse of the frequency, or
The wavelength emitted is indirectly proportional to the difference in the change in the energy level. For the wavelength 278 nm the change in energy level is significantly high. Further change in energy level is indicated by 454nm light but the difference in energy level for this wavelength to be emitted is not greater than the previous one. There is a possibility that these subsystems have now very low energy which should result in wavelengths ranging from 700 to 900 nm. There is another possibility that there is some metastable subsystems in the system which may cause LASER emission.
Answer:
6.58m
Explanation:
The kinetic energy = Workdone on the roller
Workdone = Force * distance
Given
KE = Workdone = 362J
Force = 55N
Required
Distance
Substitute into the formula;
Workdone = Force * distance
362 = 55d
d = 362/55
d = 6.58m
Hence the student must push at a distance of 6.58m
