C and E Gravitational and Chemical energy
Answer:
Explanation:
a ) If x be the position of n the bright fringe on the screen , following formula holds .
x = n (λD / d) ; λ is wave length , D is screen distance and d is slit separation .
If we increase the value of λ or wave length, x will increase so central fringe along with all the fringes will shift away from the centre .
If we increase the value of D or screen distance , it will also increase x , so fringes along with central fringe will shift away from the center.
b ) Fringes ,whether bright or dark , both shift together , either towards or away from the center .
So to move the dark spots towards the center , we need to do the opposite to what we did in the first case , ie decrease the wavelength or decrease the screen distance .
Answer:
a) He found the same value of q/m for different cathode materials.
b) y = 
, c) v = 
Explanation:
In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.
b) In the part of the plates the electrons are accelerated by the electric field,
F = ma
- e E = m a
a = - (e/m) E₀
the distance traveled is
X axis
x = v₀ t
the separation of the plates is x = d
t = vo / d
Y axis
y = v_{oy} t + ½ to t²
y = ½ a t²
y =
c) In this case there is a magnetic field B₀ and the electrons have no deflection
F = - e E + e v x B
if there is no deviation F = 0
e E = e v B
v =
Nope, this is impossible because in order for a car to pass another, they the 55 mph car would have to be behind the 65 mph car (meaning that starting ahead of the 65 mph doesn't count as pass). Assuming that they both drive for one hour, it is impossible because the first will cover a distance of 65 mi and the second would cover a distance of 55 mi. One is obviously ahead of the other and is therefore impossible for the slow one to pass the first one unless the slow car keeps driving after an hour. In that case, it would take approximately 11 minutes for it to pass the other car. This was found by finding the distance needed to pass the first car : 65 - 55 = 10 mi and converted using 1 hr/ 55 mi = .18181818 hr x 60 min/ 1 hr = 11 seconds
I hope this helps :)
If you travel 5km east then 5km west you end up back where you started therefore the vector sum would be 0.
