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Which type wire should be used to increase resistance

8_murik_8 [283]

Nichrome wire. That's the stuff that toasters are made from. The resistance is pretty high, considering the diameter. 1 meter is at about the same guage as that listed below for copper is about 96 ohms.

Most of the time you are trying to use wire with the least resistance.

A meter of copper has a listed resistance of 0.024 ohms / meter. The wire is a 19 guage wire which makes it pretty thin.

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I'm not sure what you are asking. If want the resistance of something in terms of what would increase the resistance of the same material for both calculations then

Rule 1: It you decrease the wire diameter, you increase the resistance

Rule 2: If you increase the length of the wire, you increase the resistance.

Both rules assume you are using something like copper.

7 0

3 years ago

Read 2 more answers

Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas

Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

$I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}$

For 4 particles:

$I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

$I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}$

c)

$I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

8 0

3 years ago

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago

How much metabolic energy is required for a 68kg person to run at a speed of 15km/hr for 15min ?

irga5000 [103]

<span>What I have here is exactly the same problem, however, with the time changed to 19 mins:

metabolic energy = metabolic power*time = 1.150*19*60 = 1.311 kJ..corresponding to 1.311/4.186 = 313,2 Cal or kcal

If we reasonably assume a metabolic eff.cy of 20%, it means we need to assume food for 1500 Cal approx.

Just plug the value t=15min to the equation and you will surely get the correct answer.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>

8 0

3 years ago

What is the magnitude of your displacement when you follow directions that tell you to walk 100.0m north, then 25.0m East?

atroni [7]

Answer:

Displacement from the starting position is 103.21m

Explanation:

If you draw these directions, it will create the two legs of a triangle.

Using this method, you can visualize why your displacement is what it is.

Using the pythagorean theorem

${a}^{2} + {b}^{2} = {c}^{2}$