Answer:
<em>we can say that that in an increasing order, the body is less dense than the glycerine, and the glycerine is less dense than the kerosene.</em>
<em></em>
Explanation:
If the body sinks below the kerosene level to float in glycerine, then it means that the kerosene is denser than glycerine. This is because bodies will float higher in a denser fluid that a less denser fluid. Also, since the body floats in kerosene and glycerine, then the body is less dense than the kerosene and the glycerine. Finally, <em>we can say that that in an increasing order that the body is less dense than the glycerine, and the glycerine is less dense than the kerosene</em>
this is first law of thermodynamics in which meaning of each term is given as
= change in internal energy
Q = Heat given to the system
W = work done by the system
so here we will have
∆U=Q-W , which means <u>Change in internal energy</u> in a system is equal to <u>heat given</u> added, minus <u>work</u> done by the system. the units for these variables are in <u>joule</u> .
<span>The star is a ball of hot gases
containing different kinds of elements at different cores. It has a very high
temperature that radiates all throughout the galaxy. The star has four
main parts; the core, photosphere, chromospheres and corona. The outer core of
a star located at the chromospheres contains mostly of hydrogen. Inside the hydrogen
is helium then carbon, oxygen, neon, magnesium silicon and the inert gas. When a
star releases energy it is due to the thermonuclear fusion of hydrogen into
helium. So the answer is fusion.</span>
<span>a. We can find the velocity when the camera hits the ground.
v^2 = (v0)^2 + 2ay = 0 + 2ay
v = sqrt{ 2ay }
v = sqrt{ (2)(3.7 m/s^2)(239 m) }
v = 42 m/s
The camera hits the ground with a velocity of 42 m/s
b. We can find the time it takes for the camera to hit the ground.
y = (1/2) a t^2
t^2 = 2y / a
t = sqrt{ 2y / a }
t = sqrt{ (2)(239 m) / 3.7 m/s^2 }
t = 11.4 seconds
It takes 11.4 seconds for the camera to hit the ground.</span>
The test tube will be subject to centripetal acceleration. This acceleration is given by the following formula
(accel.) = (tangential velocity)^2 / (radius)
The velocity of the probe at a distance of 10 cm from the center of the centrifuge, can be calculated using the circumference of the circle:
where omega denotes the angular velocity (radians per second). So, combining both:
The test tube is subjected to an acceleration of 18434 m/s^2!