Answer:
The differences that will be observed are;
1) The Sun will become faint and will no longer be yellow but rather appear white and will no longer be visible (become invisible) by unassisted vision as we can see the Sun today
2) The size of the Sun will shrink to a size comparable to the size of the Earth
3) The Sun will cool down and will no longer radiate as much heat
4) The nuclear reactions that generate energy on the Sun's will seize and the and the heat from the Sun will be from residual thermal energy
5) The core, which is the hottest part of the Sun will no longer be hydrogen but carbon and oxygen
Explanation:
As per the question the color of laser light is given as red.
If we arrange all the electromagnetic waves in the decreasing order of frequency ,then the electromagnetic spectrum contains gamma ray as the first which is followed by all other electromagnetic waves according to their frequency.
The visible light ranges from 400 nm to 700 nm which contains sunlight i.e white colors with it's constituent colors starting from violet to red. The red color is the longest wavelength part of the visible region.
The wavelength of visible light is longer than ultraviolet wave but smaller than infrared radiation. Except the bisible region,the color of radiation is invisible to eye.
As per the question the color of emiited laser radiation is red .Hence it must lie in the visible region of the electromagnetic spectrum.
-- As far as we know, the forces on the wheelbarrow are balanced.
-- That tells us that the net force on the wheelbarrow is zero, just
as if there were no forces acting on it at all.
-- That tells us that the wheelbarrow's acceleration is zero ... its
speed and direction of motion are not changing.
-- That tells us that the wheelbarrow is moving in a straight line
at a constant speed. It's very possible that relative to us, the speed
may be zero, but we can't tell that from the given information.
Answer:
A. the left half becomes neutral while the right half remains negatively charged
Explanation:
This is because wherever light strikes the photoconductor, it transforms from an insulator into a conductor. The charge will then migrate through it and leaves its surface. By exposing the left half of the photoconductor to light, you allow its local charge to leave and it becomes neutral.
