Answer:
Explanation:
Let T be the tension in the cord.
Impulse by cord = change in momentum of block A .
T x 5s = 10 ( 2 -0) = 20
T = 4 poundal .
acceleration of block B = 2 / 5 = 0.4 m /s²
Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .
= 8 ( 32 + .4 ) = 259.2 poundal
Frictional force on block A = 259.2 - 4 = 255.2 poundal
μ x 10 x 32 = 255.2
320μ = 255.2
μ =0 .8 .
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The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
To find the answer, we need to know about the magnetic field inside the solenoid.
<h3>What's the expression of magnetic field inside a solenoid?</h3>
- Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
- n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
- Here, n = 290/32cm or 290/0.32 = 906
I= 0.3 A
- So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.
Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
Learn more about the magnetic field inside the solenoid here:
brainly.com/question/22814970
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Answer:
ωf = 8.8 rad/s
v = 2.2 m/s
Explanation:
We will use the third equation of motion to find the maximum angular velocity of the wheel:

where,
α = angular acceleration = 6 rad/s²
θ = angular displacemnt = 1 rev = 2π rad
ωf = max. final angular velocity = ?
ωi = initial angular velocity = 1.5 rad/s
Therefore,

<u>ωf = 8.8 rad/s</u>
Now, for linear velocity:
v = rω = (0.25 m)(8.8 rad/s)
<u>v = 2.2 m/s</u>