Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r =
--------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v =
-------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v = ![\frac{1.6 * 10^{-19} * 0.3 * 0.2} {1.67*10^{-27} }](https://tex.z-dn.net/?f=%5Cfrac%7B1.6%20%2A%2010%5E%7B-19%7D%20%2A%200.3%20%2A%200.2%7D%20%7B1.67%2A10%5E%7B-27%7D%20%7D)
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s
There
are five layers of the atmosphere and these are; troposphere, stratosphere,
mesosphere, thermosphere and exosphere. The first layer, troposphere, is where
we are able to do most of our activities. This is where we can see the formation
of clouds, the production of rain, hail, snow and other weather phenomenon.
Also, this layer is where the greatest amount of air pressure because most of
the molecules of air are in this area. Like us, air has also mass and the
pressure is brought down by the earth’s gravity causing an increase in weight
exerted on you as you descend lower into the atmosphere. So, as you enter into
the other layers of atmosphere above the troposphere, the air pressure starts
to decrease. <span>Below the atmosphere
is the hydrosphere. This is where all liquid forms are located. And since the
seawater has a greater mass than air, it has the greatest pressure. </span>
i THINK question 18 is (b) and 19 are (A)
Answer:
Δx=vt−21at2
Explanation:
We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.
In this problem, the target unknown is the acceleration as the cyclist slows down.
Assuming the initial direction of travel is the positive direction, our known variables are
t=0.5s
v=10km/h
Δx=2m
Since we don't know the initial velocity (Vo), and we are not asked to find (Vo) we could use the kinematic formula that is missing (Vo) to solve for the target unknown, A
Δx=vt−21at