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andreev551 [17]
3 years ago
14

A grinding wheel with a moment of inertia of 2 kg-m2 has a 2.50 N-m torque applied to it. What is its final kinetic energy 10 se

conds after starting from rest?
312 J


237 J


156 J


106 J
Physics
1 answer:
grigory [225]3 years ago
6 0
First you do 2.5 n-m / 2 kg-m2
This equals 1.25 rad/s^2

Then multiply rad/s^2 by 10 because of the amount of seconds
This equals 12.5 rad/s^2

Then do this following equation : (1/2)Iwf^2
Do (0.5) 12.5 * 12.5 * 2
This equals 156.25 J (meaning that the answer is C: 156 J)
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Answer:

 F_Balance = 46.6 N    ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

          ∑ F = 0

          Fe –W + F_Balance = 0

         F_Balance = - Fe + W

           

The electric force is given by Coulomb's law

          Fe = k q₁ q₂ / r₂

The weight is

          W = mg

Let's replace

           F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

          q₁ = + 8 μC = +8 10⁻⁶ C

          q₂ = - 3 μC = - 3 10⁻⁶ C

          r = 0.3 m = 0.3 m

Let's calculate

         F_Balance = 5 9.8 - 8.99 10⁹  8 10⁻⁶ 3 10⁻⁶ / (0.3)²

         F_Balance = 49 - 2,397

         F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

          m' = F_Balance / g

          m' = 46.6 /9.8

          m' = 4,755 kg

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