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3 years ago

7

There is a 120 V circuit in a house that is a dedicated line for the dishwasher, meaning the dishwasher is the only resistor on

that circuit line. If the dishwasher draws 18 A of electricity, what would the resistance of the dishwasher be? Round the answer to the nearest hundredth of an ohm. Ω

2 answers:

Katena32 [7]3 years ago

8 0

R = V/I

R = 120v/18A

R = 6.67 Ohms

... NO MATTER how many other devices are connected (in parallel) on the same circuit.

Fantom [35]3 years ago

5 0

Answer:

The answer is 6.67 Ohms

Explanation:

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A current I flows down a wire of radius a.

Helga [31]

Answer:

(a) $K = \frac{I}{2\pi a}$

(b) $J = \frac{I}{2\pi as}$

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

$K = \frac{dI}{dL}$

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

$dL = 2\pi r$

The radius of the wire = a

$dL = 2\pi a$

The surface current density $K = \frac{I}{2\pi a}$

(b) The current density is inversely proportional

$J \alpha s^{-1}$

$J = \frac{k}{s}$ ......(1)

k is the constant of proportionality

$I = \int\limits {J} \, dS$

$I = J \int\limits \, dS$ ........(2)

substituting (1) into (2)

$I = \frac{k}{s} \int\limits\, dS$

$I = k \int\limits^a_0 \frac{1}{s} {s} \, dS$

$I = 2\pi k\int\limits\, dS$

$I = 2\pi ka$

$k = \frac{I}{2\pi a}$

substitute $J = \frac{k}{s}$

$J = \frac{I}{2\pi as}$

7 0

2 years ago

Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o

Dominik [7]

Answer:

The center of mass for the object is $y_c = 1.063 \ m$ from the origin

Explanation:

From the question we are told that

The mass of the first object is $m_1 = 1.99 \ kg$

The position of first object with respect to origin $y_1 = 2.99 \ m$

The mass of the second object is $m_2 = 2.96 \ kg$

The position of second object with respect to origin $y_2 = 2.57 \ m$

The mass of the third object is $m_3 = 2.43 \ kg$

The position of third object with respect to origin $y_3 = 0 \ m$

The mass of the fourth object is $m_3 = 3.96 \ kg$

The position of fourth object with respect to origin $y_3 = -0.502 \ m$

Generally the center of mass of the object along the x-axis is zero because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

$y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}$

=> $y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96 + 2.43 + 3.96}$

=> $y_c = 1.063 \ m$

3 0

3 years ago

One student did an experiment on the rock cycle.

Nonamiya [84]

First, when the student added the layers of wax over each other, this became a representation of sedimentary rocks.

Then the student folded his/her palm and squeezed the layers of wax. This means that the student applied heat and pressure on the wax (sedimentary rocks)

Referring to the diagram below which represents the rock cycle, we will find that applying heat and pressure on sedimentary rocks would convert these rocks into metamorphic rocks.

Based on the above, the best choice would be:
<span>d. Heat and pressure can change sedimentary rocks into metamorphic rocks.</span>

7 0

3 years ago

Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be

Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂² → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² % = 0,036 %

4 0

3 years ago

Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it conta

shtirl [24]

Answer:

1020 km

Explanation:

A complete rotation of the wheel equals a distance of 1 circumference.

The circumference is

$C = \pi d$

where <em>d</em> is the diameter of the wheel.

300,000 rotations = $300000\pi d = 300000\times\pi\times1.08\text{ m} = 1017876.0\ldots\text{ m}$

In kilometers, this is = 1017876/1000 km = 1020 km

6 0

3 years ago