Explanation:
n=50,r=0.02mn=50,r=0.02m,
I=5AandB=0.20TI=5AandB=0.20T
τismaxiμmwhensinθ=90∘τismaxiμmwhensinθ=90∘
τmax=niabsin90∘=mbτmax=niabsin90∘=mb
=50×5×3.14×4×10−4×2×10−1=50×5×3.14×4×10-4×2×10-1
=6.28×10−2Nm=6.28×10-2Nm
Given τ=12×τmaxτ=12×τmax
⇒sinsinθ=12⇒sinsinθ=12 or sinθ=30∘sinθ=30∘
=∠betweenareavar→randmag≠ticfield=∠betweenareavar→randmag≠ticfield.
So angle between magnetic field and the plane of the coil
=90∘−30∘=60∘=90∘-30∘=60∘.
<h3>HOPE IT HELPS </h3>
<h2>mark me in brainliest answers please please please </h2>
When resistance force on a lever increases, nothing happens automatically.
But if you want to keep lifting the load, then YOU must increase the force of
your effort in order to make it happen.
The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = (
-
) / (
+
)
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
Learn more about Equilibrium here: brainly.com/question/517289
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