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:
<h3>1.01 s</h3>
Explanation:
Using the equation of motion S = ut+1/2gt² to solve the problem where;
u is the initial velocity of the chocolate = 0m/s
t is the time taken
g is the acceleration due to gravity = 9.81m/s²
S is the height of fall = 5.0m
Substituting the given parameter into the formula to get the time t we have;
5 = 0(t)+1/2(9.81)t²
5 = 4.905t²
t² = 5/4.905
t² = 1.019
t = √1.019
t = 1.009 secs
<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>
Answer:81.235N
Explanation:
Work=88J
theta=10°
distance=1.1 meters
work=force x cos(theta) x distance
88=force x cos10 x 1.1 cos10=0.9848
88=force x 0.9848 x 1.1
88=force x 1.08328
Divide both sides by 1.08328
88/1.08328=(force x 1.08328)/1.08328
81.235=force
Force=81.235
<h2>
Answer: 13.61 N/m</h2>
Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force
applied to it, <u>as long as the spring is not permanently deformed</u>:
(1)
Where:
is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.
is the length of the spring without applying force.
is the length of the spring with the force applied.
According to this, we have a spring where only the force due gravity is applied.
In other words, the force applied is the weigth
of the block:
(2)
Where
is the mass of the block and
is the gravity acceleration.
(3)
(4)
Knowing the force applied
and
and
, we can substitute the values in equation (1) and find
:
(5)
(6)
<u>Finally:</u>