Answer:
Explanation:
The force exerted in a magnetic field is given as
F = q (v × B)
Where
F is the force entered
q is the charge
v is the velocity
B is the magnetic field
Given that,
The magnetic field is
B = 2•i + 4•j. T
The velocity of the electron is
v = 2•i + 6•j + 8•k. m/s
Also, the charge of an electron is
q = -1.602 × 10^-19 C.
Then note that,
V×B is the cross product of the speed and the magnetic field
Then,
F = q (V×B)
F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)
Note
i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]
F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]
F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]
F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)
F = -1.602 × 10^-19(16•j - 32•i)
F = -1.602 × 10^-19 × ( -32•i + 16•j)
F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j
Then, the x component of the force is
Fx = 5.126 × 10^-18 N
Also, the y component of the force is
Fy = -2.563 × 10^-18 N
Answer:
210
Explanation:
A ball rolls horizontally off the cliff at a speed of 30 m/s. It takes 7 seconds for the ball to hit the ground. What is the height of the cliff and the horizontal distance traveled by the ball?
S = (1/2)*9.8 m/s^2 * 7^2 = 240.1 m if the ball is very dense so air resistance, and therefore terminal velocity, can be ignored.
S = v * t = 30 m/s * 7 s = 210 m for the horizontal distance, again assuming negligible air resistance.
<span>The gravity of earth depends on the magnetism from its core. as this magnetism increases, the magnitude of the gravity increases.</span>
Answer:
The angular velocity is
5.64rad/s
Explanation:
This problem bothers on curvilinear motion
The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s
We know that the velocity v is given as
v= ωr
Where ω is the angular velocity
r is 300mm to meter = 0.3m
the radius of the circle
described by the level
v=1.64m/s
Making ω subject of the formula and solving we have
ω=v/r
ω=1.64/0.3
ω=5.46 rad/s
Answer:
Coefficient of friction = 0.836
Explanation:
If v be the speed after one quarter of the circular path
v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8
tangential acceleration = 5.8 m/s²
radial acceleration = v² /r = 5.8
total acceleration = √2 x 5.8
m x√2 x 5.8 = m x g xμ
μ = √2 x 5.8 / 9.8 = 0.836
