F=K*X,
F=M*a
M*a=K*X
2.5*9.81=K*0.0276
24.525=K*0.0276
24.525/0.0276=K
K= 888.6 N/m ---- force constant
assuming 2.5 refers to the new extension, just divide F/ 0.025
to get
981N/m
Please state the options and I will answer to the best of my abilities XD
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
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.
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