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kipiarov [429]
3 years ago
5

A toaster oven has a resistive heating element. The average rate at which it dissipates energy as thermal energy is 1.00 kW. In

the United States, emf amplitude in household circuits is Emax = 170 V and the AC oscillation rate is 60 Hz.
What is the root-mean-square current through the heating element?
Physics
1 answer:
Mandarinka [93]3 years ago
7 0

Answer:

I = 8.31 A

Explanation:

given,

Thermal Energy = 1 kW

Emf amplitude = 170 V

AC oscillation = 60 Hz

E_{rms} = \dfrac{E_{max}}{\sqrt{2}}

E_{rms} = \dfrac{170}{\sqrt{2}}

E_{rms} =120.21\ V

Rms current is calculated as

P = VI

I =\dfrac{P}{V}

I =\dfrac{1000}{120.21}  

I = 8.31 A

the root-mean-square current through the heating element is equal to I = 8.31 A

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Kelvin is a base unit of temperature scale from SI that defines as zero degree Kelvin (absolute zero). The absolute zero is a hypothetical statement that all molecular movement stops because there is no transient of energy for the molecules to move. When converting temperature in degree Celsius to Kelvin, add 273. You are given 600K and you are asked to find it in degrees Celsius.  

T(K) = T(C) + 273
600 K = T(C) + 273
T(C) = 600 – 273
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Which of the following are relatively unchanged fragments from the early period of planet building in the solar system? a. aster
Inga [223]

Answer:

The answer would be B ........m

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4 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
4 years ago
An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).
Anvisha [2.4K]

Answer:

denisity = 52.33 g/cm^{3}

Explanation:

Density:

d = \frac{m}{v}

We have that m = 785 and that v = 15 cm^{3}.

d = \frac{785}{15}

d = 52.33 m^{3}

4 0
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Un caballo parte del reposo y alcanza una velocidad de 15m/s en un tiempo de 8s. ¿ Cuál fue su aceleración y que distancia recor
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Answer: A 120 metros por segundo

Explanation: multiplicas la velocidad por el tiempo

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