All categories

1 year ago

In some cases fixture wires may be used for

1 answer:

3 0

You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.

<h3>What are some uses of fixture wires?</h3>

Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG

In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.

To know more about fixture wires, refer

brainly.com/question/26098282

#SPJ4

You might be interested in

A 25 kg box is being pulled at a constant velocity with a tension force of 65 N. what is the coefficient of friction between the

DiKsa [7]

The box is in equilibrium, so Newton's second law says

<em>n</em> + (-<em>w</em>) = 0

65 N + (-<em>f</em> ) = 0

where <em>n</em> denotes the magnitude of the normal force, <em>w</em> denotes the weight of the box, and <em>f</em> denotes the magnitude of the friction force.

The box has a weight of

<em>w</em> = (25 kg) (9.80 m/s²) = 245 N

so <em>n</em> = 245 N, too.

The friction force has magnitude

<em>f</em> = 65 N

and is proportional to the normal force by a factor of <em>µ</em>, the coefficient of kinetic friction. So we have

65 N = <em>µ</em> (245 N) → <em>µ</em> ≈ 0.26

7 0

3 years ago

Which coefficient is needed in front of NaNO3 to balance the equation Na2S + Zn(NO3)2 → ZnS + _NaNO3? 1, 2, 5, 7,

maksim [4K]

The coefficient is needed in front of NaNO3 to balance the equation is 2.

<h3>What is balanced equation?</h3>

When in a chemical reaction, both the product elements or compounds should have number of moles equal to that of the elements or compounds of reactants.

In the given chemical reaction equation,

Na2S + Zn(NO3)2 → ZnS + _NaNO3

Number of moles of S is 1 on both the product and reactant side. Zn has 1 mole on both side. NO3 has 2 mole and Na has 2 mole on reactant side. So to balance the equation, Na and NO3 both must have 2 moles on product side.

Thus, the coefficient is needed in front of NaNO3 to balance the equation is 2.

Learn more about balanced equation.

brainly.com/question/11322377

#SPJ1

6 0

2 years ago

The rotating blade of a blender turns with constant angular acceleration of 1.51 rad/s^2. How much time does it take to reach an

yawa3891 [41]

time to reach an angular velocity of 36.8 is 24.370 s.

<h3></h3><h3>What is angular acceleration?</h3>

The temporal rate at which angular velocity changes is referred to as angular acceleration. Naturally, there are two forms of angular acceleration, referred to as spin angular acceleration and orbital angular acceleration, just as there are two types of angular velocity, namely spin angular velocity and orbital angular velocity. As opposed to orbital angular acceleration, which is the angular acceleration of a point particle around a fixed origin, spin angular acceleration describes the angular acceleration of a rigid body about its centre of rotation.

w(t) = w(0) + α*t

also w(0) =0

=> time = 36.8/1.51= 24.370 s

to learn more about angular acceleration go to - brainly.com/question/21278452

#SPJ4

6 0

2 years ago

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in

valkas [14]

Answer:

Volume of the sample: approximately $\rm 0.6422 \; L = 6.422 \times 10^{-4} \; m^{3}$ .

Average density of the sample: approximately $\rm 2.77\; g \cdot cm^{3} = 2.778 \times 10^{3}\; kg \cdot m^{3}$ .

Assumption:

$\rm g = 9.81\; N \cdot kg^{-1}$ .
$\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ .
Volume of the cord is negligible.

Explanation:

<h3>Total volume of the sample</h3>

The size of the buoyant force is equal to $\rm 17.50 - 11.20 = 6.30\; N$ .

That's also equal to the weight (weight, $m \cdot g$ ) of water that the object displaces. To find the mass of water displaced from its weight, divide weight with $g$ .

$\displaystyle m = \frac{m\cdot g}{g} = \rm \frac{6.30\; N}{9.81\; N \cdot kg^{-1}} \approx 0.642\; kg$ .

Assume that the density of water is $\rho(\text{water}) = \rm 1.000\times 10^{3}\; kg \cdot m^{-3}$ . To the volume of water displaced from its mass, divide mass with density $\rho(\text{water})$ .

$\displaystyle V(\text{water displaced}) = \frac{m}{\rho} = \rm \frac{0.642\; kg}{1.000\times 10^{3}\; kg \cdot m^{-3}} \approx 6.42201 \times 10^{-4}\; m^{3}$ .

Assume that the volume of the cord is negligible. Since the sample is fully-immersed in water, its volume should be the same as the volume of water it displaces.

$V(\text{sample}) = V(\text{water displaced}) \approx \rm 6.422\times 10^{-4}\; m^{3}$ .

<h3>Average Density of the sample</h3>

Average density is equal to mass over volume.

To find the mass of the sample from its weight, divide with $g$ .

$\displaystyle m = \frac{m \cdot g}{g} = \rm \frac{17.50\; N}{9.81\; N \cdot kg^{-1}} \approx 1.78389 \; kg$ .

The volume of the sample is found in the previous part.

Divide mass with volume to find the average density.

$\displaystyle \rho(\text{sample, average}) = \frac{m}{V} = \rm \frac{1.78389\; kg}{6.42201 \times 10^{-4}\; m^{3}} \approx 2.778\; kg \cdot m^{-3}$ .