Answer:
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Explanation:
blah blah blah your welcome
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:
11:1
Explanation:
At constant acceleration, an object's position is:
y = y₀ + v₀ t + ½ at²
Given y₀ = 0, v₀ = u, and a = -g:
y = u t − ½g t²
After 6 seconds, the ball reaches the maximum height (v = 0).
v = at + v₀
0 = (-g)(6) + u
u = 6g
Substituting:
y = 6g t − ½g t²
The displacement between t=0 and t=1 is:
Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]
Δy = 6g − ½g
Δy = 5½g
The displacement between t=6 and t=7 is:
Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]
Δy = (42g − 24½g) − (36g − 18g)
Δy = 17½g − 18g
Δy = -½g
So the ratio of the distances traveled is:
(5½g) / (½g)
11 / 1
The ratio is 11:1.
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
