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3 years ago

Which type of resistor is commonly used in automotive circuits? a. Fixed value b. Stepped c. Variable d. All of the choices

Physics

1 answer:

soldi70 [24.7K]3 years ago

7 0

Answer:d-All of the above

Explanation:

A resistor is a device used in electrical circuits that offers resistance to current flow.In automotive circuit commonly used resistor are

Fixed value
Stepped
Variable

Fixed value:A fixed value electrical resistance is one whose value is resistance value is fixed and does not change in circuit.

Stepped resistance:Stepped resistor also called as tapped resistors provide two or more set resistor values. In a circuit, the various resistances are attached to different terminals. When the switch is shifted, different resistance values are put in the circuit.

Variable resistance:The variable resistor gives more control over the current flow by increasing the resistance. When the resistance of the variable resistor increases, the amount of current allowed to flow in the circuit decreases.

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Which of two factors influence the weight of an object due to gravitational pull?

densk [106]

The weight of an object when calculated by multiplying with the pull of the gravity is dependent on the mass of the object and the value of g. The value of g is constant however is still dependent on the distance of the object from the center of the Earth. Thus, the answers are <em>mass and distance. </em>

5 0

3 years ago

Can someone help me?!!!!!

german

<h2>Hello!</h2>

The answer is:

The first option, the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

$Distance=NorthCoveredDistance+EastCoveredDistance$

$Distance=4*180m+3*180m=720m+540m=1260m$

Actual distance between the start and the end point (displacement):

$ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m$

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

$DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m$

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

3 0

3 years ago

An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.

Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis $E=31.25\ N/c$ and $E = 15.625\ N/c$ respectively.

Explanation:

Given :

Distance between charges , $d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.$

Magnitude of charges , $q=1\ nC = 10^{-9}\ C.$

Dipole moment , $p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.$

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

$E=\dfrac{2kp}{r^3}$

Putting all values and $r=12\times 10^{-2}\ m.$

We get , $E=31.25\ N/c.$

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

$E = \dfrac{kp}{r^3}$

Putting all values and $r=12\times 10^{-2}\ m.$

We get , $E = 15.625\ N/c.$

Hence , this is the required solution.

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3 years ago

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bearhunter [10]

Answer:

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Explanation:

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masya89 [10]

Answer:

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Explanation:

See attachment for calculations

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2 years ago

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