Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
Answer:
Bar graph
Explanation:
each day collects data so a bar graph would work.
Answer:
3.0 x 10¹ Nm
Explanation:
Torque = F x r
Where F is force applied and r is perpendicular distance from pivot point . r
is also called lever arm
Here F = 15 N and r = 2.0 m
Torque
= 15 N X 2.0 m
= 3.0 10¹ Nm.
Answer:
a = 7.5 m / s²
Explanation:
For this exercise let's use Newton's second law, let's create a coordinate system with the x axis parallel to the plane and the y axis perpendicular to the plane
Y axis
N - W cos θ = 0
N = mg cos θ
X axis
W sin θ = m a
mg sin θ = m a
a = g sin θ
let's calculate
a = 9.8 cos 40
a = 7.5 m / s²
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