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Aleonysh [2.5K]
3 years ago
13

A toaster is plugged into a 132 volt household circuit it draws the amps of current If you use a toaster 2 hours a week how much

energy do you use in kWh
Physics
1 answer:
Ne4ueva [31]3 years ago
8 0
There is a missing data in the text of the problem: the current drawn is 0.5 A (found on internet)

Solution:
The power used by the toaster is the product between the voltage V and the current I:
P=VI=(132 V)(0.5 A)=66 W=0.066 kW

The toaster is used for a time of 2 hours: t=2h, therefore the energy used is equal to
E=Pt =(0.066 kW)(2 h)=0.132 kWh
You might be interested in
Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T
Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

T-f-T'=m_Ba

Where m_B=Mass of block B

Substitute the values

50-10-T'=m_B\times 0=0

T'==40 N

For block A

T'-f'=m_Aa

Where m_A=Mass of block A

Substitute the values

40-f'=m_A\times 0=0

f'=40 N

Hence, the friction force acting on block A=40 N

3 0
3 years ago
1) A train accelerates from 36 km/hr to 54 km/hr in 10<br> s. Find acceleration?
Crank

Answer: Given:

Initial velocity= 36km/h=36x5/18=10m/s

Final velocity =54km/h=54x5/18=15m/s

Time =10sec

Acceleration = v-u/ t

=15-10/10=5/10=1/2=0.5 m/s2

Distance =s=?

From second equation of motion:

S=ut +1/2 at^2

=10*10+1/2*0.5*10*10

=100+25

=125m

So distance travelled 125m

Hope it helps you

3 0
3 years ago
Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believe
KIM [24]

Answer:

- the expected value is 8

- the standard deviation is 2.8284

Explanation:

Given the data in the question;

The model N(t), the number of planets found up to time t, as a poisson process,

∴ N(t) has distribution of poisson distribution with parameter (λt)

so

the mean is;

λ = 1 every month = 1/3 per month

E[N(t)] = λt

E[N(t)] = (1/3)(24)

E[N(t)] = 8

Therefore, the expected value is 8

For poisson process, Variance and mean are the same,

Var[N(t)] = Var[N(24)]

Var[N(t)] = E[N(24)]

Var[N(t)] = 8

so the standard deviation will be;

σ[N(24)] = √(Var[N(t)] )

σ[N(24)] = √(8 )

σ[N(24)] = 2.8284

Therefore, the standard deviation is 2.8284

8 0
3 years ago
Which of the objects below has the greatest acceleration? *
lesya692 [45]

\\ \bull\sf\dashrightarrow F=ma

\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}

Now

\\ \bull\sf\dashrightarrow a\propto\dfrac{1}{m}

\\ \bull\sf\dashrightarrow a\propto F

  • Lower mass=Higher acceleration
  • Lower Force=Lower Acceleration

Option B has lowest mass and highest force hence its correct

8 0
2 years ago
Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation
Artemon [7]

Answer:

L =   μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

        E_{L} = - L dI / dt

 

Let's use Faraday's law

       E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

        - d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

        n d Ф_B = L dI

we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

       L =   μ₀ n r / 2I

4 0
3 years ago
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