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3 years ago

7

At a cost of 9.0 cents/kWh, estimate how much this would add to your monthly electric energy bill if you made toast four morning

s per week (16 times per month).

1 answer:

vredina [299]3 years ago

3 0

The question is incomplete, here is the complete question:

How many kWh of energy does a 650-W toaster use in the morning if it is in operation for a total of 6.0min ?

At a cost of 9.0 cents/kWh, estimate how much this would add to your monthly electric energy bill if you made toast four mornings per week (16 times per month).

Express your answer using two significant figures.

Answer:

$ 0.1 per month

Explanation:

We convert from Watt to kilowatt

650/1000 = 0.65KW

We convert from minutes to hour:

6/60= 0.1 hr

Convert cents to dollars

9/100 = $0.09

Energy cost in KWh= 0.65 × 0.1 × 0.09 ×16= $ 0.1 per month

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Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

$Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }$

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

$Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

Sum of the potential at point p is

V = Va + Vb

that is

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

$\frac{1}{(a^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }$

likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

$V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

by reciprocal rule

1/a² = a⁻²

$V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]$

by binomial expansion of fractional powers

where $(1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...$

if we expand the expression we have the equivalent as shown

${(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )$

also,

${(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )$

the above equation becomes

$V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]$

$V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )$

$V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )$

Answer

$V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }$

OR

$V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }$

8 0

3 years ago

Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.0095 m

PIT_PIT [208]

Answer:

Current in outer circle will be 15.826 A

Explanation:

We have given number of turns in inner coil $N_I=170$

Radius of inner circle $r_i=0.0095m$

Current in the inner circle $I_i=8.9A$

Number of turns in outer circle $N_o=150$

Radius of outer circle $r_o=0.015m$

We have to find the current in outer circle so that net magnetic field will zero

For net magnetic field current must be in opposite direction as in inner circle

We know that magnetic field is given due to circular coil is given by

$B=\frac{N\mu _0I}{2r}$

For net magnetic field zero

$\frac{N_I\mu _0I_I}{2r_I}=\frac{N_O\mu _0I_0}{2r_O}$

So $\frac{170\times \mu _0\times 8.9}{2\times 0.0095}=\frac{150\times \mu _0I_O}{2\times 0.015}$

$I_O=15.92A$

4 0

2 years ago

How much power is needed to lift the 200-N object to a height of 10 m in 4 s?

harkovskaia [24]

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000 Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

Power = 2000 / 4 Nm/s = 500 Nm/s = 500 watts

6 0

2 years ago

How do the positions of the sun and moon affect what people do?

Verdich [7]

-- The position of the sun was originally the primary influence in determining
when people went to sleep and when they woke up. Although it no longer
directly influences us, that pattern is so deeply ingrained in our make-up
that our behavior still largely coincides with the positions of the sun.

-- The position of the Moon was originally the primary influence in determining
the cycle of human female physiology. Although it no longer directly influences
us, that pattern is so deeply ingrained in human make-up that the female cycle
still largely coincides with the positions of the Moon.

6 0

3 years ago

A ball rolls for 8 seconds and travels 24 meters. How fast was it traveling?

belka [17]

Answer:

The speed of the ball was, v = 3 m/s

Explanation:

Given data,

The time period of the ball, t = 8 s

The distance the ball rolled, d = 24 m

The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,

v = d / t m/s

Substituting the given values in the above equation,

v = 24 / 8

= 3 m/s

Hence, the speed of the ball was, v = 3 m/s

8 0

3 years ago