Answer:
E- The star becomes a red giant (LATEST STAGE)
F- The surface of the star becomes brighter and cooler
C- Pressure from the star's hydrogen-burning shell causes the non burning envelope to expand
A- The shell of hydrogen surrounding the star's nonburning helium core ignites.
D- The star's non burning helium core starts to contract and heat up
B- Pressure in the star's core decreases (EARLIEST STAGE)
(A star moves away from the main sequence once its core runs out of hydrogen to fuse into helium. The energy once supplied by hydrogen burning reduces and the core starts to compress under the force of gravity. This contraction allows the core and surrounding layers to heat up. Finally, the hydrogen shell around the core becomes hot enough to ignite hydrogen burning.
Answer:
It can be seen from the operation of pin-hole camera, formation of shadows and eclipse.
Explanation:
The phenomenon of light traveling in a straight line is known as rectilinear propagation of light.
One this evidence can be seen from the operation of pin-hole camera, which depends on rectilinear propagation of light
Also two natural effects that result from the rectilinear propagation of light are the formation of Shadows and Eclipse.
Answer:
3.62m/s and 2.83m/s
Explanation:
Apply conservation of momentum
For vertical component,
Pfy = Piy
m* Vof (sin38) - m*Vgf (sin52) = 0
Divide through by m
Vof(sin38) - Vgf(sin52) = 0
Vof(sin38) = Vgf(sin52)
Vof (sin38/sin52) = Vgf
0.7813Vof = Vgf
For horizontal component
Pxf= Pxi
m* Vof (cos38) - m*Vgf (cos52) = m*4.6
Divide through by m
Vof(cos38) + Vgf(cos52) = 4.6
Recall that
0.7813Vof = Vgf
Vof(cos38) + 0.7813 Vof(cos52) = 4.6
0.7880Vof + 0.4810Vof = 4.
1.269Vof = 4.6
Vof = 4.6/1.269
Vof = 3.62m/s
Recall that
0.7813Vof = Vgf
Vgf = 0.7813 * 3.62
Vgf = 2.83m/s
Answer:
Closest to the dog.
Explanation:
Sounds are louder the closer you are to them.
