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tankabanditka [31]

3 years ago

14

How many places are there for electrons in the innermost shell of an atom?

1 answer:

kumpel [21]3 years ago

7 0

The innermost shell of an atom is the shell corresponding to the principal quantum number of 1. Because the principal quantum number is 1, the angular quantum number and the magnetic quantum number of that shell must be 0. Therefore, only 2 electrons can possibly be in the innermost shell, one with spin up and the other with spin down.

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tester [92]

Answer:

choose to spend limited resources to meet their needs

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3 years ago

A material that can easily flow is called a

murzikaleks [220]

A material that can easily flow is called a...

A. Fluid.

3 0

3 years ago

Read 2 more answers

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Fofino [41]

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 \times 10^{-2})(782.6)$

$a_t = 24.3 m/s^2$

6 0

3 years ago

Hot water is poured into a mug and the mug gets hot. This is an example of which type of energy transfer?

Korolek [52]

The answer is B: Conduction

5 0

2 years ago

Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at 600 K, and heat is supplied from the dome at a rate

SIZIF [17.4K]

Answer:

$\epsilon_2=0.098$

Explanation:

Diameter $d=30cm=0.3m$

Temperature $T=600k$

Rate of supply $r=65W$

Emissivity of base surface $\in_b =0.55$

Temperature at base $T_b=400k$

Generally the equation for Area of base surface is mathematically given by

$A_b=\frac{\pi}{4}d^2$

$A_b=\frac{\pi}{4}0.3^2$

$A_b=0.0707m^2$

Generally the equation for Area of Hemispherical dome is mathematically given by

$A_h=\frac{\pi}{2}d^2$

$A_h=\frac{\pi}{2}0.3^2$

$A_h=0.1414m^2$

Since base is a flat surface

$F_{11}+F_{12}=1$

$F_{11}=0$

Therefore

$F_{12}=1$

$A_b=0.0707m^2$

Generally the equation for Net rate of radiation heat transfer between two surfaces is mathematically given by

$Q_{21}=-Q_{12}$

$Q_{21}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\epsilon}{A_b\epsilon_1} +\frac{1}{A_bF_{12}} +\frac{1-\epsilon_2}{A_h*\epsilon_2} }$

Where

$\sigma=5.67*10^{-8}$

Therefore

$65=\frac{(5.67*10^{-8}(400^4-600^4))}{\frac{1-0.55}{0.0707*0.55}+\frac{1}{0.0707}+\frac{1-\epsilon_2}{0.1414*\epsilon_2}}$

$\epsilon_2=0.098$

$\epsilon_2 \approx 0.1$

Therefore the emissivity of the dome is

$\epsilon_2=0.098$

3 0

3 years ago