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gregori [183]
3 years ago
10

A mass m is attached to a string connected to a force sensor on a rotating platform. The platform’s angular velocity, ω, can be

easily measured, but the linear velocity of the mass cannot. When the mass is at r = 0.20 m and rotates at a constant ω = 8.5 rad/s, a force sensor reads 4.8 N. What is the mass m?
Physics
1 answer:
makkiz [27]3 years ago
7 0

Answer:

The mass m is 0.332 kg or 332 gm

Explanation:

Given

The platform is rotating with angular speed , \omega =8.5\, \frac{rad}{sec}

Mass m is moving on platform in a circle with radius , r=0.20\, m

Force sensor reading to which spring is attached , F=4.8\, N

Now for the mass m to move in circle the required centripetal force is given by F=m\omega ^{2}r

=>4.8=m\times 8.5 ^{2}\times 0.20

=>m=0.332\, kg

Thus the mass m is 0.332 kg or 332 gm

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Answer:

The distance that separates the two particles is 7.42 cm.

Explanation:

Given;

the mass of each particle, m = 3 mg = 3 x 10⁻⁶ kg

the magnitude of charge of each particle, q = 6.0 nC

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F = ma = \frac{kq^2}{r^2}

a = v/t

where;

a is the acceleration of the two particles

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t is time

v = u + gt

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5 = 9.8t

t = 5/9.8

t = 0.51 s

a = v/t

a = 5/0.51 = 9.8 m/s²

Total force on the two particles = (2m)a = (2* 3 x 10⁻⁶)9.8

F = 5.88 x 10⁻⁵ N

Substitute in the value of F in the above equation and calculate r

F = \frac{kq^2}{r^2}

where;

k is coulomb's constant = 8.99 x 10⁹ Nm²/c²

r is the distance of separation between the two particles

F = \frac{kq^2}{r^2} \\\\r^2 = \frac{kq^2}{F}\\\\r = \sqrt{ \frac{kq^2}{F}} = \sqrt{ \frac{8.99*10^9*(6*10^{-9})^2}{5.88*10^{-5}}} = 0.0742 \ m

Therefore, the distance that separates the two particles at the instant when each has a speed of 5.0 m/s, is 7.42 cm.

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