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3 years ago

The most abundant element in earth's continental crust (by weight) is____?

Physics

2 answers:

sashaice [31]3 years ago

8 0

The answer is A. Oxygen

AlekseyPX3 years ago

5 0

Answer: A. Oxygen

Explanation: Oxygen is the most abundant element in the Earth's crust by weight. Oxygen amounts to 46.6% of the Earth's crust by weight.

Oxygen is a major compound of the silicate minerals where it reacts with other elements. Oxygen can he will receive bottling collected pls, he will pretence as a compound in carbonates and phosphates. Oxygen has industrial, medical, and commercial purposes. It is used with acetylene to cut and weld metals. It is used in hospitals to ease respiratory disease and can also be used to manufacture explosives among other numerous uses.

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what is a point of view of an object used to determine another obejects motion i nedd help asap plsss

Igoryamba

A believe that’s called a reference point.

9 0

4 years ago

Later research indicated that electric current is actually the flow of ____

iren [92.7K]

Answer:

Electrons.

Explanation:

Electricity was discovered before the discovery of electrons by J.J Thompson in 1896. Before the electron, it was thought that it is the positive ions that move through the wire and carry current—that's why today the conventional current represents the flow of positive charges.

After J.J Thompson's discovery of the electrons, it was realized that it is the electrons that actually carry the current through the conductor. But changing the direction of the conventional current didn't seem appropriate, and that's why the convention continues to be used to this day—reminding us that once it were the positive ions that were thought to carry the current.

6 0

3 years ago

Which equation represents mass-energy equivalence? E = m2c E = mc2 E = (mc)2 E = mc

motikmotik

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

$E = mc^2$

where

m = mass in kg

c = speed of light in m/s

this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle

So correct relation above will be

$E = mc^2$

4 0

3 years ago

Read 2 more answers

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

3 years ago

Read 2 more answers

If an automobile had a 100%-efficient engine, transferring all of the fuel's energy to work, would the engine be warm to your to

svetlana [45]

Answer:

The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.

Explanation:

In this ideal engine, none of these events would happen due to the nature of the efficiency.

We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.

n=W, total/(E, available).

However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.

5 0

3 years ago