Answer:
the magnitude of the magnetic force on the wire is 0.2298 N
Explanation:
Given the data in the question;
we know that, the magnitude of magnetic force is given as;
|F | = I(
×
)
given that
I = 2.6 A
= 0.17
= 0.52
so we substitute
|F | = 2.6( 0.17i" × 0.52j" )
|F | = 0.2298 N
Therefore, the magnitude of the magnetic force on the wire is 0.2298 N
Answer:
= -32.53 m / s
this velocity is directed downwards
Explanation:
This is a free fall exercise, let's use the expression
= v_{oy}^{2} + 2 g (y -yo)
where we are assuming that there is friction with the air, as the body falls its initial velocity is zero
v_{oy} = √ 2g (y - y₀)
let's calculate
v_{y} = √ (2 9.8 (0-54.0))
= -32.53 m / s
this velocity is directed downwards
The mass on the spring is 0.86 kg
Explanation:
The period of a mass-spring system is given by the equation
where
m is the mass
k is the spring constant
In this problem, we have:
k = 88.7 N/m is the spring constant
The system makes 15 oscillations in 9.24 s: therefore, the period of the system is
Now we can re-arrange the first equation to solve for the mass:
Learn more about period:
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