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Rzqust [24]
3 years ago
9

One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i

s, it promotes the warming of earth's atmosphere. calculate the annual production of carbon dioxide in kilograms if there are exactly 40.0 million cars in the united states and each car covers a distance of 6990 mi at a consumption rate of 21.4 miles per gallon. enter your answer in scientific notation.
Physics
1 answer:
Musya8 [376]3 years ago
4 0

The annual production of carbon dioxide is 124121.49×10^{6}[/tex] kg.

First we calculate the fuel consumed by each car in a year

Fuel consumed=6990/21.4=326.63 gallon

Now we calculate the amount of fuel consumed by 40 million cars in a year

Fuel consumed=326.63*40*10^6=13065.42 million gallon,

Now we can calculate the annual production of carbon dioxide in the USA

CO2 production rate=9.50*13065.42=124121.49*10^6 kg

Therefore the annual production of carbon dioxide in USA is 124121.49×10^{6}[/tex] kg

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A simple generator is used to generate a peak output voltage of 19.0 V . The square armature consists of windings that are 6.65
salantis [7]

One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,

\epsilon = NBA\omega

Where,

N = Number of loops

B = Magnetic Field

A = Cross-sectional Area

\omega = Angular velocity

Re-arrange to find N,

N = \frac{\epsilon}{BA\omega}

Our values are given as,

\epsilon = 19V

B = 0.434T

\omega = 49.8\frac{rev}{s} (\frac{2\pi rad}{1 rev}) = 99.6\pi rad/s

A = (6.65*10^{-2})^2 m^2

Replacing at our equation we have:

N = \frac{\epsilon}{( 0.434)A\omega}

N = \frac{19}{( 0.434)((6.65*10^{-2})^2)(99.6\pi)}

N = 31.63 \approx 32

Therefore the number of loops of wire should be wound on the square armature is 32 loops

6 0
3 years ago
Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_{1} +q_{2} =-q_{3} -----eqution 1

where,

q_{1} is the heat absorbed by the solid at 0⁰C

q_{2} is the heat absorbed by the liquid at 0⁰C

q_{3} the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature

Again,

q=n*\delta {_f_u_s} -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_{2} = -q_{3}

<u>Step 4:</u> calculate the final temperature of the water

79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}

Substitute in the values; we will have,

79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})

79.13 kJ + 990.66J* (T_{f}-218}) = -1463J*(T_{f}-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_{f}-218}) = -1.463KJ*(T_{f}-100})

79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3

collect like terms,

2.45366T_{f} = 283.133

∴T_{f} = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0
3 years ago
How does the nitrogen enter the food web?
nikklg [1K]

When bacteria in the soil takes nitrogen from the air it becomes nitrates it can finally move through the food chain in this form.

6 0
3 years ago
An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50
Setler79 [48]

Answer:

very hard others will answer it

Explanation:

hard

6 0
2 years ago
In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs
bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

V=\frac{I}{nAq}

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

\frac{I}{q} = 1.2*10^{18}

A= 1.3*10^{-8}m^2

n=6.3*10^{28} e/m^3

\omicron{O} = 1.2*10^{-4}(m/s)(N/c) Mobility

We can find the drift velocity replacing,

V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}

V= 1.465*10^-3m/s

The electric field is given by,

E= \frac{V}{\omicron{O}}

E=\frac{1.465*10^-3}{1.2*10^{-4}}

E=12.2V/m

7 0
3 years ago
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