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Novay_Z [31]
3 years ago
13

The SAME amount of current I passes through three different resistors. R2 has twice the cross-sectional area and the same length

as R1, and R3 is three times as long as R1 but has the same cross-sectional area as R1. 1)In which case is the CURRENT DENSITY through the resistor the smallest
Physics
1 answer:
Snezhnost [94]3 years ago
5 0

Answer:

resistor R₂ has the lowest current density

Explanation:

The current density is

          j = I / A

now let's analyze each case

a) R₂ has an area 2A₀ and a length L₀ that R₁

b) R₃ has an area Ao and a length 3L₀ what R₁

we can see that all the area is given in relation to the resistance R₁

 

the current density in R₁ is

         j₁ = I / A₀

the current density in R₂

         j₂ = I / 2A₀

         j₂ 2 = ½ I/A₀

the current density in R₃

         j₃ = I / A₀

         j₂ < j₁ = j₃

therefore resistor R₂ has the lowest current density

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Consider a system made of a ramp that has a run of 8 m and a rise of 6 m. Two equal masses of 4-kg are connected to each other b
djyliett [7]

Answer:

The force of friction is 15.68 N.

Explanation:

Given:

Rise is 6 m and run is 8 m.

Mass of each body (m) = 4 kg

The whole system is in equilibrium.

Now, consider the diagram below representing the scenario given in the question.

The forces acting on the mass that is hanging are tension force in the upward direction and weight of the body acting vertically downward.

As the mass is in equilibrium, the total upward force equals total downward force. So,

T= mg=4\times 9.8=39.2\ N -------- (1)

Now, the forces acting on the other mass along the ramp are:

1. Tension (T) up the ramp

2. mg sin\theta and frictional force (f) down the ramp

Now, as per question:

Rise = 6 m and run = 8 m

So, from figure,

\tan\theta=\frac{6}{8}=0.75\\\\\theta=\tan^{-1}(0.75)=37\°

Now, \sin\theta=\sin(37\°)=0.6

Now, as the other mass is also at rest, net force acting on it is also 0. So,

F_{net}=0\\\\T-(mg\sin\theta+f)=0\\\\mg\sin\theta+f=T\\\\f=T-mg\sin\theta

Now, plug in the given values and solve for 'f'. This gives,

f=39.2-(4\times 9.8\times 0.6)\\\\f=39.2-23.52=15.68\ N

Therefore, the force of friction is 15.68 N

4 0
4 years ago
When a new path of lesser resistance is made for an existing circuit a(n) _____________ circuit occurs.
Margaret [11]

When a new path of lesser resistance is made for an existing circuit a(n) short circuit occurs.

<h3>What is short circuit?</h3>

An electrical circuit short circuit is when two nodes that are supposed to be connected at different voltages make an improper connection. This leads to an electric current that can damage circuits, cause overheating, fire, or explosions, and is only constrained by the network's remaining nodes' equivalent Thevenin resistance. While short circuits are typically the result of a failure, they can occasionally be brought on purpose, such as when voltage-sensing crowbar circuit protectors are being installed.

An electrical connection that requires two nodes to have the same voltage is known as a short circuit in circuit analysis. Since there is no resistance and hence no voltage drop across the link in a "perfect" short circuit, there is no short circuit.

To learn more about short circuit, visit:

brainly.com/question/13260673

#SPJ4

5 0
2 years ago
How to find the coefficient of kinetic friction?
ycow [4]
It depends on the type of question, mechanical condition and given values, 
You can use the formula, 

Coefficient of kinetic friction  =  Force of Kinetic friction  /  Normal force (perpendicular to contacting surfaces)

Hope this helps!
5 0
3 years ago
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim
Nonamiya [84]

Answer:

The distance covered by puck A before collision is  z = 8.56 \ m

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  v_A =  3.90 \ m/s

        The speed of puck B is  v_B  =  4.30 \ m/s

The distance covered by puck A is mathematically represented as

     z =  v_A * t

  =>  t  =  \frac{z}{v_A}

 The distance covered by puck B  is  mathematically represented as

      18 - z =  v_B  * t

=>   t  = \frac{18 - z}{v_B}

Since the time take before collision is the same

        \frac{18 - z}{V_B}  =  \frac{z}{v_A}

substituting values

          \frac{18 -z }{4.3}  = \frac{z}{3.90}

=>      70.2 - 3.90 z   = 4.3 z

=>       z = 8.56 \ m

8 0
3 years ago
How do machines increase mechanical advantage?
Kryger [21]
Mechanical advantage is a measure of the force amplification achieved by using a tool.
7 0
3 years ago
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