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saw5 [17]
3 years ago
15

A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for

gas heating and cooking in a period of 1 month. Given that 1.00 therm is equal to 29.329.3 kWh, how many milligrams of mass would need to be converted directly to energy each month to meet the energy needs for the home?
Physics
1 answer:
faltersainse [42]3 years ago
5 0

Answer:

Explanation:

Given that,.

A house hold power consumption is

475 KWh

Gas used is

135 thermal gas for month

Given that, 1 thermal = 29.3 KWh

Then,

135 thermal = 135 × 29.3 = 3955.5 KWh

So, total power used is

P = 475 + 3955.5

P =4430.5 KWh

Since 1 hr = 3600 seconds

So, the energy consumed for 1hr is

1KW = 1000W

P = energy / time

Energy = Power × time

E = 4430.5 KWhr × 1000W / KW × 3600s / hr

E = 1.595 × 10^10 J

So, using Albert Einstein relativity equation

E = mc²

m = E / c²

c is speed of light = 3 × 10^8 m/s

m = 1.595 × 10^10 / (3 × 10^8)²

m = 1.77 × 10^-7 kg

Then,

1 kg = 10^6 mg

m = 1.77 × 10^-7 kg × 10^6 mg / kg

m = 0.177mg

m ≈ 0.18 mg

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An object will not change its motion unless an outside force acts on it true or false
Gre4nikov [31]

Answer:

True!

Explanation:

Newton's First Law :)

7 0
2 years ago
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In an HVAC system, the purpose of pneumatics is to serve as A. a controller B. conditional air flow C. an energy source D. a sou
Naya [18.7K]

Answer: <u><em>A</em></u>

Explanation:

A pneumatic control system uses compressed air as a method of control for HVAC systems. ... Each senor responds to changes in temperature, humidity, and static pressure as examples, to provide feedback in a control loop to open or close the actuator to meet the control set point.

3 0
3 years ago
A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c
Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current densityJ=2.8\times10^{7}\ A/m^2

We need to calculate the current

Using formula of current density

J = \dfrac{I}{A}

I=J\timesA

Where, J = current density

A = area

I = current

Put the value into the formula

I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2

I=1.97=2.0\ A

Hence, The current is 2.0 A.

7 0
3 years ago
A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘
hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of  mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution,T_f = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC

\Delta T_f=i\times K_f\times m

Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}

5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})

On solving we get:

\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of Ca(NO_3)_2 in the mixture:

\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%

The mass percentage of calcium nitrate is 31.23%.

5 0
3 years ago
A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me
loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

Q = mC_p \Delta T

Where,

m = mass

C_p= Specific Heat

\Delta T = Change in temperature

Replacing with our values we have that

m = 0.1g

C_p = 139J/Kg\cdot K \rightarrow Specific heat of mercury

\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K

Replacing

Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J

Therefore the heat lost by mercury is 0.09J

5 0
3 years ago
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