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elixir [45]
3 years ago
9

How many kWh of energy does a 550-W toaster use in the morning if it is in operation for a total of 5.0 min? At a cost of 9.0 ce

nts/k Wh, estimate how much this would add to your monthly electric energy bill if you made toast four mornings per week.
Physics
1 answer:
iren [92.7K]3 years ago
3 0

Answer:

0.0458 kWh

6.5736 cents

Explanation:

The formula for electric energy is given as

E = Pt................. Equation 1

Where E = Electric energy, P = Electric power, t = time.

Given; P = 550 W, t = 5 min = (5/60) h = 0.083 h.

Substituting into equation 1

E = 550(0.083)

E = 45.83 Wh

E = (45.83/1000) kWh

E = 0.0458 kWh.

Hence the kWh = 0.0458 kWh.

If the makes a toast four morning per week, and the are Four weeks in a month.

Total number days he makes toast in a month = 4×4 = 16 days.

t = 16×0.083 h = 1.328 h.

Total energy used in a month = 550(1.328)

E = 730.4 Wh

E = 0.7304 kWh.

If the cost of energy is 9.0 cents per kWh,

Then for 0.7304 kWh  the cost will be 9.0(0.7304) = 6.5736 cents.

Hence this would add 6.5736 cents to his monthly electric bill

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You observe lighting striking and then hear to sound 8 seconds later. The speed of sound in air is 340 m/s. How
Leto [7]

Answer:

2.72 Kilometers

Explanation:

8 × 340 m/s = 2720 m = 2.72 Kilometers

7 0
2 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
2 years ago
Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How fa
BartSMP [9]

Answer

given,

Echo time = t = 1000 s

speed = ?

v = \dfrac{d}{t}

a) speed of electromagnetic wave

c = \dfrac{d}{t}

wave travels Venus two time

c = \dfrac{2d}{t}

d = \dfrac{ct}{2}

d = \dfrac{3 \times 10^8\times 1000}{2}

d = 1.5 x 10¹¹ m

b) now, echo time

    c = \dfrac{2d}{t}

    t = \dfrac{2d}{c}

    t = \dfrac{2\times 75}{3 \times 10^8}

          t = 5 x 10⁻⁷ s

8 0
3 years ago
Which statement best describes energy and matter in a closed system? (2 po
Ivahew [28]

Answer:

Energy can flow into and out of the system but matter cannot.

Explanation:

In a closed system, energy can flow in and out of the system but matter cannot.

  • A closed system prevents double way flow of matter.
  • A closed system conserves matter.

For an isolated system, energy and matter cannot flow out of the system.

For open systems, energy and matter can flow out of the system.

Such systems are used for certain thermodynamics experiment.

4 0
3 years ago
You have a flag pole at an angle of 45 degrees to a wall it is designed to support a max torque of 40Nm. You bought a new one wi
In-s [12.5K]

Answer:

0.915 rad or 52.44 degrees

Explanation:

Let g = 10 m/s2. We can calculate the gravity force acting on center of mass of the new 8.75 kg pole:

F = mg = 8.75 * 10 = 87.5 kg

The moment is the dot product of force and moment arm

M = \vec{F} \cdot \vec{r} = Frcos\theta = 87.5*0.75cos\theta = 65.625cos\theta

where θ is the angle between the gravity force and the pole, or between the pole and the wall. As M can be at its max, which is 40 Nm, we can solve for θ:

62.625cos\theta = 40

cos\theta = 40 / 62.625 = 0.61

\theta = cos^{-1}0.61 = 0.915 rad = 0.915 * 180 /\pi = 52.44^o

7 0
3 years ago
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