They protested by marching in the streets.
Answer:
Bone
Explanation:
Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.
The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.
The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.
Answer: Tides are periodic rises and falls of large bodies of water. Tides are caused by gravitational interaction between the earth and the moon. The gravitational attraction of the moon causes the oceans to bulge out in the direction of the moon.
Answer:
84.82N/C.
Explanation:
The x-components of the electric field cancel; therefore, we only care about the y-components.
The y-component of the differential electric field at the center is
.
Now, let us call
the charge per unit length, then we know that
;
therefore,


Integrating

![$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$](https://tex.z-dn.net/?f=%24E%20%3D%20%5Cfrac%7Bk%20%5Clambda%20%20%20%7D%7BR%7D%2A%5B-cos%28%5Cpi%20%29%2Bcos%280%29%20%5D%24)

Now, we know that


and the radius of the semicircle is

therefore,


Mass, m = 5890g
Change in temperature, θ = Final_temperature - Initial_temperature
= 315 - 462°C
= -147°C
Specific heat capacity of aluminum, c = 0.900 J/(g*K)
=mcθ
=5890g x 0.900 J/(g*K) x -147°C
=-779,247j
Answer would be C.