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2 years ago

The principal quantum number is given the symbol and has positive, whole-number values starting from?

Physics

1 answer:

deff fn [24]2 years ago

6 0

Answer:

Principal quantum numbers are denoted by the symbol 'n'. They designate the principal electron shell of the atom. The positive whole number values starting from 1,2,3 and so on..

Explanation:

All orbitals that have the same value of n are said to be in the same shell (level). For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state. The total number of orbitals for a given n value is n2.

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A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

ivann1987 [24]

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

$\tau=F\times r$

$\tau=82.0\times0.150$

$\tau=12.3\ N-m$

Now, The angular acceleration

$\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}$

$\dfrac{d\omega}{dt}=20.73\ rad/s^2$

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

$\tau=I\times\dfrac{d\omega}{dt}$

$I=\dfrac{I}{\dfrac{d\omega}{dt}}$

$I=\dfrac{12.3}{20.73}$

$I= 0.593\ kg-m^2$

Hence, The moment of inertia of the wheel is 0.593 kg-m².

5 0

3 years ago

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

Fiesta28 [93]

Answer:

a) $P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

b) For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

Explanation:

For this case we have the following data given:

$D= 1mm = 0.001 m$ represent the diameter

$r = D/2= 0.0005m$ represent the radius

$T= 5.7 N$ represent the tension

$f = 57 Hz$ represent the frequency of the oscillator

$A= 0.54 cm = 0.0054 m$ represent the amplitude of the wave

Part a

For this case we can assume that the power transmitted to the wave is the same power of the oscillator. and we have the following formula for the power:

$P= 2 \pi^2 \rho S v f^3 A^2$

This expression can be written in different ways:

$P= 2 \pi^2 \rho S \sqrt{\frac{T}{\mu}} f^2 A^2$

$P= 2\pi^2 \rho S \sqrt{\frac{T}{\rho S}} f^2 A^2$

$P= 2 \pi^2 f^2 A^2 \sqrt{S \rho T}$

Where f is the frequency , T the tension $rho= 7800 \frac{kg}{m^3}$ the density of the steel, A the amplitude and $S= \pi r^2$ the area, so then we have everuthing in order to replace and we got:

$P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

Part b

For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

8 0

4 years ago

When are objects in thermal equilibrium? *

frutty [35]

Thermal equilibrium is achieved when two objects or systems reach the same temperature and cease to exchange energy through heat. When two objects are placed together, the object with more heat energy will lose that energy to the object with less heat energy.

7 0

3 years ago

If a spring is stretched 4m from its starting length when 20n of force is applied, then how much work (in joules) is done by the

lys-0071 [83]

ANSWER:

250 J

STEP-BY-STEP EXPLANATION:

F = 20N is required to stretch the spring by 4 meters

We know that the force is equal to:

$F=k\cdot x$

We solve for k (spring constant):

$k=\frac{F}{x}=\frac{20}{4}=5\text{ N/m}$

The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:

$\begin{gathered} W=\frac{1}{2}k\cdot x^2 \\ \text{ Replacing} \\ W=\frac{1}{2}\cdot5\cdot10^2 \\ W=250\text{ J} \end{gathered}$

The work required is 250 joules.

5 0

2 years ago

A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects a beat frequency of 700 Hz. The speed of

ASHA 777 [7]

Answer:

3.948m/s

Explanation:

To solve this we need to apply Doppler effect theory

To find the frequency received by insect will be gotten when the Source and observer both are moving in same direction which is given by

f1 = f0 x (V - Vo)/(V - Vs)

f0 = 30.0 kHz

V = 344 m/s

Vs will now be the speed of the bat and

Vo will be the speed of the object which is = 0 m/s

So substituting we have

f1 = 30 x 10^3 x (344- 0)/(344- Vs)

Next to find the frequency reflected by wall we use

f2 = f1 x (V + Vs)/(V + Vo)

So substituting the value of f1 calculated above we have

f2 = 30 x 10^3 x (344 + Vs) x (344 - 0)/[(344 - Vs) x (344 + 0)]

f2 = 30 x 10^3 x (344 + Vs)/(344- Vs)

But the beat frequency detected by bat is 700 Hz,

So we say

f2 - f0 = 700 Hz

30 x 10^3 x (344+ Vs)/(344 - Vs) - 30x 10^3 = 700

(344 + Vs)/(344 - Vs) = 1 + 700/30000 = 1.023

344 + Vs = 344 x 1.023 - Vs x 1.0233

Vs = 344 x ( 1.023 - 1)/(1 + 1.023)

So finally

Vs = Speed of source that is the bat is = 3.949m/s

6 0

4 years ago