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Nataliya [291]
3 years ago
12

A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula

r to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk?
Physics
1 answer:
Marina86 [1]3 years ago
3 0

Answer:\alpha =10.66 rad/s^2

Explanation:

Given

mass of disk m=5 kg

diameter of disc d=30 cm

Force applied F=4 N

Now this force will Produce  a  torque of magnitude

T=F\cdot r

T=4\dot 0.15

T=0.6 N-m

And Torque is given Product of moment of inertia and angular acceleration (\alpha )

T=I\cdot \alpha

Moment of inertia for Disc I= \frac{Mr^2}{2}

I=0.05625 kg-m^2

0.6=0.05625\cdot \alpha

\alpha =10.66 rad/s^2

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Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge
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Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

h) 24.3 N θ = 44.2º

Explanation:

a) As the electric force is exerted along the line that joins the charges, due to any of the charges A or C has non-zero x-coordinates, the force has no x components either.

So, Fcax = 0

b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

Fyca = 29. 9 N

c) In order to get the magnitude of the force exerted by B on C, we need to know first the distance between both charges:

rbc² = (3.00 m)² + (4.00m)² = 25.0 m²

⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

cos θ = x/r = 4.00 / 5.00 m =

Fcbx = 21.7*(-0.8) = -17.4 N

e) The  y component can be calculated in the same way, projecting the force over the y-axis, as follows:

Fcby = Fcb* sin (-θ) = 21.7* (-3.00/5.00) = -13.0 N

f) The sum of both x components gives :

Fcx = 0 + (-17.4 N) = -17.4 N

g) The sum of both y components gives :

Fcy = 29.9 N + (-13.0 N) = 16.9 N

h) The magnitude of the resultant electric force acting on C, can be found just applying Pythagorean Theorem, as follows:

Fc = √(Fcx)²+(Fcy)² = (17.4)² + (16.9)²\sqrt{((17.4)^{2} +(16.9)^{2}} = 24.3 N

The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

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Oxygen-18 is a naturally-occuring, stable isotope and is commonly used is scientific studies as a tracer. Using the periodic tab
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Oxygen has Atomic number 8 so all isotopes have 8 protons and 8 electrons.

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3 years ago
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Answer:

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