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Answer:
angular acceleration is -0.2063 rad/s²
Explanation:
Given data
mass m = 95.2 kg
radius r = 0.399 m
turning ω = 93 rpm
radial force N = 19.6 N
kinetic coefficient of friction μ = 0.2
to find out
angular acceleration
solution
we know frictional force that is = radial force × kinetic coefficient of friction
frictional force = 19.6 × 0.2
frictional force = 3.92 N
and
we know moment of inertia that is
γ = I ×α = frictional force × r
so
γ = 1/2 mr²α
α = -2f /mr
α = -2(3.92) /95.2 (0.399)
α = - 7.84 / 37.9848 = -0.2063
so angular acceleration is -0.2063 rad/s²
You need to provide a picture or tell us the examples... we can’t see what you see
So we need to find the formula for magnetic field B using the current (I) and the distance from the probe (d). So, We know that the stronger the current I, the stronger the magnetic field B. That tells us that the I and B are proportional. Also we know that the strength of the magnetic field B is weaker as the distance d of the probe increases. That tells us that B and d are inversely proportional. So our formula should have B=(I/d)*c where c is a constant of proportionality. c=μ₀/2π where μ₀ is the permeability of free space. So finally our formula is B=(μ₀I)/(2πd).
Answer
given,
length of bar = 80 cm
mass of the bar = 10 kg
smaller mass = 4 kg
distance = 20 cm
taking moment about B
difference between two scale = 8 - 6
= 2 N
