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Answer:
the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)
the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)
Explanation:
Use Biot, Savart, the magnetic field
Given that,
i = 1.00A
d → l = 4.00 m m ^ j
r = 2.5m
Displacement vector is
=2.5m
on the axis of x at x = 2.5
r = 2.5m
And unit vector
Therefore, the magnetic field is as follow
(Along z direction)
B)r = 5.00m
Displacement vector is
=5.00m
on the axis of x at x = 5.0
r = 5.00m
And unit vector
Therefore, the magnetic field is as follow
(Along x direction)
Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
<u>α = - 1.047 rad/s²</u>
B.
We can use the second equation of motion to find the angular distance:
<u>θ = 14.1 rad</u>
C.
θ = (14.1 rad)(1 rev/2π rad)
<u>θ = 2.24 rev</u>
Answer:
v = 7.67 m/s for L= 1m
Explanation:
Let's use the conservation of mechanical energy, at the highest point and the lowest point
Initial. Vertical ruler
Em₀ = mg h
Final. Just before touching the floor
= K = ½ I w²
Em₀ =
m g h = ½ I w²
The moment of inertia of a ruler that turns on one end is
I = 1/3 m L²
Let's replace
m g h = ½ (1/3 m L²) w²2
g h = 1/6 L² w²
They ask for the speed of the end so the height h is equal to the length of the ruler
g L = 1/6 L² w²
The linear and angular variables are related
v = w r
w = v / r
In this case the point of interest a in strangers r = L
g L = 1/6 L² v² / L²
v = √ 6 g L
Let's calculate
Assume that the length of the meter is L = 1 m
v = √ (6 9.8 1)
v = 7.67 m/s