Answer:
C) upward
Explanation:
The problem can be solved by using the right-hand rule.
First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).
Now we can apply the right hand rule to the charged particle:
- index finger: velocity of the particle, to the right
- middle finger: direction of the magnetic field, out of the page
- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward
Therefore, the direction of the magnetic force is upward.
Velocity of an object is its rate of change of the object's position per interval of time. Velocity is a vector quantity which means that it consists of a magnitude and a direction. Magnitude is represented by the speed and the direction is represented by the angle. To determine the velocity components, we use trigonometric functions to determine the angle of the components. For the north component we, use the sine function while, for the west component, we use the cosine function. We calculate as follows:
north velocity component = (16.8 m/s) (sin 54°) = 16.4 m/s
<span>west velocity component = (16.8 m/s) (cos 54°) = 3.49 m/s</span>
Answer:
Explanation:
We know that
Acceleration due to gravity g is given by the formula
G= gravitational constant
M= mass of the Earth
r= radius of the earth
Let acc. due to gravity after radius is 20% greater be g_2
then
=> g1/g2 = (r_2/r_1)^2 => g2 = 9.81/1.2^2 = 6.8125
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
