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agasfer [191]
2 years ago
12

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.
Physics
2 answers:
aliya0001 [1]2 years ago
7 0

The info given in the question:

Voltage= 120V

Current=18A

Now we have to find the resistance. To find it use the following formula:

V=IR

Now making R to be the subject of the formula

R=V/I

R=120/18

The answer is 6.67 ohms

As dishwasher is the only resistor in the line the voltage drop is going to be 120V. The resistance values determines the hindrance that is present in the circuit that opposes the free flowing electrons  

Papessa [141]2 years ago
7 0

Answer : The resistance of the dishwasher is R=6.67\ A.

Explanation :

It is given that in a house there is only one resistor on the circuit line. There is a 120 V circuit in a house that is a dedicated line for the dishwasher. The dishwasher draws the current of 18 A of electricity.

Using Ohm's law in this circuit such that, V = I R

Where R is the resistance offered to the circuit.

So, using V and I in the equation of Ohm's law we get :

R=\dfrac{V}{I}

R=6.67\ A

Hence, this is the required solution.

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An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
If a single constant force acts on an object that moves on a straight line, the object's velocity is a linear function of time.
olya-2409 [2.1K]

Answer:

F=mkv

Explanation:

Given that

v = v_i - kx

We know that acceleration a given as

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v = v_i - kx

\dfrac{dv}{dt}=\dfrac{dv_i}{dt}-k\dfrac{dx}{dt}

\dfrac{dv}{dt}=0-k\dfrac{dx}{dt}

We know that

F=m\dfrac{dv}{dt}

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F=-mkv

So the magnitude of force F

F=mkv

5 0
2 years ago
A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate
liq [111]

Answer:

3 cm

Explanation:

According to the question,

D=1.5 m.

d=9 cm.

\lambda =0.9 mm.

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

y=2\frac{\lambda (D)}{d}.

Substitute all the variable in above equation.

y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}.

y=3 cm.

5 0
2 years ago
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
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Answer:

(a) A = 3.90 \AA

(b) A = 4.50 \AA

(c) A = 5.51 \AA

(d) A = 9.02 \AA

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As per the question:

Radius of atom, r = 1.95 \AA = 1.95\times 10^{- 10} m

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = 2\times 1.95 = 3.90 \AA

(b) For body centered cubic lattice:

A = \frac{4}{\sqrt{3}}r

A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA

(c) For face centered cubic lattice:

A = 2{\sqrt{2}}r

A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA

(d) For diamond lattice:

A = 2\times \frac{4}{\sqrt{3}}r

A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA

6 0
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ch4aika [34]

Answer:

4.41 W

Explanation:

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Given that P = 0.0625 when V = 1.50:

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So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

P = (12.6)² / 36

P = 4.41

So the power consumption is 4.41 W.

5 0
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