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3 years ago

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What picture of the electron is suggested by the quantum-mechanical model of the hydrogen atom? a. a raisin in pudding c. a plan

etary orbiting body b. a probability cloud d. a light quantum

1 answer:

Ivanshal [37]3 years ago

3 0

D. a light quantum this is the answer .

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The charges on the proton and electron are typically measured in: Select the correct answer below: volts amperes coulombs dalton

malfutka [58]

Answer: columbs

Explanation:

Electrical charge are measured in columbs, usually demoted as C. Hence, the charges on proton and electron will be measured in Coloumbs. It typically measures the amount of electricity conveyed per second by a current of 1 ampere. The other units Given such as ; Volt is used for measuring voltage, which is the pressure in an electrical source. AMPERE is used for measuring the current flowing through an electrical circuit.

Dalton is a unit of mass and is about 1.660 * 10^-27 kg

4 0

2 years ago

Which objects in space formed from the huge disk cisce and debris beyond the outer planets? Select two option

Dennis_Churaev [7]

Answer:

Comets

Explanation:

The Kuiper Belt is a collection of trans-Neptunian objects that consist of comets and other dwarf planets, including Pluto.

4 0

3 years ago

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

What two kinds of crust are involved in a subduction zone

Anarel [89]

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

5 0

3 years ago

(a) The volume flow rate in an artery supplying the brain is 3.4 x 10-6 m3/s. If the radius of the artery is 5.6 mm, determine t

UNO [17]

Answer: a) speed = 3.45 × 10^-2 m/s

b) speed = 1.38 × 10^-1 m/s

Explanation: shown in the attachment

3 0

3 years ago