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Greeley [361]
2 years ago
9

What are the two principal parts of an AC induction motor? 

Physics
1 answer:
damaskus [11]2 years ago
3 0
The two principal parts of an AC induction motor are the stator and the rotator.


Hope this helped! (:
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50 J of work was performed in 20 seconds. How much power was used to do this task?
yuradex [85]
Power=work/time
power=50/20
50/20=2.5
Therefore A. 2.5 W
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2 years ago
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Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc
jarptica [38.1K]
Calculate the weight of the table through the equation,

   W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
   W = (0.44 kg)(9.8 m/s²) 
   <em>W = 4.312 N</em>

The components of this weight can be calculated through the equation,

   Wx = W(sin θ) 

and Wy = W(cos θ)

x - component:
   Wx = W(sin θ)
Substituting,
  Wx = (4.312 N)(sin 150°) = <em>2.156 N</em>

  Wy = (4.312 N)(cos 150°) =<em> -3.734 N</em>
6 0
2 years ago
Two identical carts travel at the same speed toward each other, and then a collision occurs. The graphs show the momentum of eac
madam [21]

Explanation :

The interaction between two objects is termed as the collision. The collision can be of two types i.e. elastic collision and inelastic collision.

In this case, two identical carts travel at the same speed toward each other, and then a collision occurs. In an inelastic collision, the momentum before and after the collision remains the same but its kinetic energy gets lost.

After the collision, both the object sticks over each other and moves with one velocity.

Out of the given graph, the graph that shows a perfectly inelastic collision is attached. It shows that after the collision both the carts move with the same velocity.

5 0
2 years ago
Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father
-Dominant- [34]
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
6 0
2 years ago
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In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following pressure p,
viktelen [127]

Answer:

Explanation:

In case of gas , work done

W = ∫ p dV , p is pressure and dV is small change in volume

the limit of integration is from Vi to Vf .

= ∫ p dV

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= p₀ V^{-\frac{6}{5} +1} / ( \frac{-6}{5} +1 )

=  - 5p₀ V^{-\frac{1}{5}

Taking limit from Vi  to Vf

W = - 5 p₀ ( V_f^\frac{-1}{5} - V_i^{\frac{-1}{5}  ) ltr- atm.

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