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:
Potential energy is 
Explanation:
The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.
Potential energy
In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.
But we know the values of g and h
So, the potential energy
m is the mass of the hiker, wich is not in the description of the problem.
The instant it was dropped, the ball had zero speed.
After falling for 1 second, its speed was 9.8 m/s straight down (gravity).
Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.
Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.
ANYTHING you drop does that, if air resistance doesn't hold it back.
Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
