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vekshin1
3 years ago
5

A high school physics student is sitting in a seat reading this question. The magnitude of the force with which the seat is push

ing up on the student to support him is closest to what?
Physics
2 answers:
liberstina [14]3 years ago
7 0

Answer:

Force due to seat must be equal to the Weight of the student

Explanation:

As we know that the student is sitting on the seat at rest

So here we can say that at this position the student must be at equilibrium position

so we can write force equation for this position of student

vertically downward the net force is due to his/ her weight which is given as

W = mg

now this weight of the student is counter balanced by the the normal force due to seat which is in opposite direction

Now for equilibrium position of the object we will have

F_n = mg

so the seat will push the student upwards by force which is equal to the weight.

Dennis_Churaev [7]3 years ago
4 0
The force pushing down  is the force of Gravity. On a chair it is in perfect balance with the force pushing up (the normal force)
 

in terms of magnitude 
FN = FG = mg

the forces are in opposite direction

hope this helps
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57:23
jok3333 [9.3K]

Answer:

X: Low potential energy

Y: High Potential energy

Z: Flow of electrons

Explanation: From the figure, it's obvious that Z is the flow of electrons, as shown by the arrow demonstrating the direction of the flow. Because of this, we can easily nullify choices B and C.

From the figure, we can notice that Y has more energy stored and X has a lot less, so you can conclude that Y has high potential energy while X has low potential energy.

8 0
3 years ago
When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When
Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, \dfrac{dI}{dt}=0.0240\ A/s

Induced emf, \epsilon=12.4\ mV=12.4\times 10^{-3}\ V

Current, I = 1.5 A

Magnetic flux, \phi=0.00338\ Wb

The induced emf through the solenoid is given by :

\epsilon=L\dfrac{dI}{dt}

or

L=\dfrac{\epsilon}{(di/dt)}........(1)

The self inductance of the solenoid is given by :

L=\dfrac{N\phi}{I}.........(2)

From equation (1) and (2) we get :

\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}

N is the number of turns in the solenoid

N=\dfrac{\epsilon I}{\phi (dI/dt)}

N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0
2 years ago
How many components can be realized of a vector?​
stiv31 [10]

Answer:

<em>two different components</em>

Explanation:

<em>Any two-dimensional vector can be conceived of as having two distinct components. The component of a single vector describes the vector's effect in a specific direction.</em>

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2 years ago
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7 0
3 years ago
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Which of the following modifications to a solenoid would be most likely to decrease the strength of its magnetic field?
Goryan [66]

By reading the fine details of the question, carefully and analytically, I have determined that there's no list of modifications to choose from.

The strength of the magnetic field of a solenoid depends on the electric current in its coil windings, the number of wire turns in its coil windings, and the material in its core.

In order to <em>DE</em>crease the strength of its magnetic field, any one or more of these steps could do the job:

-- DEcrease the electric current in its coil windings.  This can be accomplished by decreasing the voltage of the power source that energizes the coil, and/or increasing the resistance of the wire in the coil.

-- DEcrease the number of wire turns in the coil.

-- If the solenoid has anything in its core, change the core to something with a lower magnetic 'permeability'.  An Iron core will produce the greatest magnetic field strength.  Air, vacuum, or NO core will produce the lowest magnetic field strength.

8 0
3 years ago
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