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>
The mass on the left has a downslope weight of
W1 = 3.5kg * 9.8m/s² * sin35º = 19.7 N
The mass on the right has a downslope weight of
W2 = 8kg * 9.8m/s² * sin35º = 45.0 N
The net is 25.3 N pulling downslope to the right.
(a) Therefore we need 25.3 N of friction force.
Ff = 25.3 N = µ(m1 + m2)gcosΘ = µ * 11.5kg * 9.8m/s² * cos35º
25.3N = µ * 92.3 N
µ = 0.274
(b) total mass is 11.5 kg, and the net force is 25.3 N, so
acceleration a = F / m = 25.3N / 11.5kg = 2.2 m/s²
tension T = 8kg * (9.8sin35 - 2.2)m/s² = 27 N
Check: T = 3.5kg * (9.8sin35 + 2.2)m/s² = 27 N √
hope this helps. :)
Answer:
Mechanical advantage = 3
Explanation:
Given:
Weight = 300 N
Force load = 100 N
Find:
Mechanical advantage:
Computation:
Mechanical advantage = Weight/Force Load
Mechanical advantage = 300/100
Mechanical advantage = 3
First let us calculate for the angle of inclination using
the sin function,
sin θ = 1 m / 4 m
θ = 14.48°
Then we calculate the work done by the movers using the
formula:
W = Fnet * d
So we must calculate for the value of Fnet first. Fnet is
force due to weight minus the frictional force.
Fnet = m g sinθ – μ m g cosθ
Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48
Fnet = 84.526 N
So the work exerted is equal to:
W = 84.526 N * 4 m
<span>W = 338.10 J</span>