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zmey [24]
3 years ago
12

;-; please help me....​

Physics
1 answer:
sineoko [7]3 years ago
5 0
The answer is 24N. Since the body is moving with constant velocity all the forces must balance (equal & opposite)
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Answer:

39 m/s

Explanation:

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Which of the following usually occurs with a short circuit? a. All parts of the circuit will begin to carry higher amounts of cu
fgiga [73]

Answer:

Hi,

The correct answer option is; D. Most of the current will flow through one part of the circuit.

Explanation:

A short circuit is a low resistance path in an electric connection between two conductors supplying current in a circuit.

It happens when excess amounts of current flow in the power source through a 'short path'.

Short circuits occur at very high temperatures which is of course caused by the heat produced during dissipation.

An example of application of short circuit is arc welding, where heating is achieved through short circuit.

6 0
3 years ago
What is the relationship between electric motors, generators, and transformers?
KengaRu [80]

Answer:

A generator turns rotary motion into electricity. It is basically the inverse of a motor. Generally a transformer changes one voltage into another based on the number of conductor windings on each side. There are two sets of windings called the “primary” and the “secondary”.

Explanation:

6 0
4 years ago
A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
3 years ago
Please help me I have been struggling
disa [49]
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4 0
3 years ago
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