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3 years ago

9. Steradian is the angel which lies in: a) One dimension b) Two dimensions

Physics

2 answers:

Marizza181 [45]3 years ago

8 0

Answer:

Explanation:

m_a_m_a [10]3 years ago

6 0

Answer:

<h2>Question:Steradian is the angel which lies in?</h2>

<h2>Answer:A.)One dimension</h2><h2 />

Explanation:

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<h2>THANK YOU!!!</h2>

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What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?

DIA [1.3K]

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

$F=kx$

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

$k=\frac{15.0-0}{10.0-0}=1.5 N/m$

Therefore when the spring has a elongation of $x=3.0 m$ , its potential energy is

$E=\frac{1}{2}(1.5)(3.0)^2=6.8 J$

Learn more about potential energy:

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3 years ago

In a hydrogen atom, the electron in orbit around the proton feels an attractive force of about 7.45 × 10−8 N. If the radius of t

Tcecarenko [31]

The attraction force provides the electron's centripetal force.

8.30^-8N = mrω² 

ω² = 8.30^-8 / (9.11^-31 kg x 4.70^-11m) .. .. ω² = 1.94^33 (rad/s)² .. .. ω = 4.40^16 rad/s 

f = ω/2π = 4.40^16 / 2π .. .. .. ►f = 7.0^15 Hz

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3 years ago

An electron is trapped in a one-dimensional infinite well of width 340 pm and is in its ground state. What are the (a) longest,

Nesterboy [21]

Answer:

(a) 1.2703×10⁻⁷ m

(b) 4.7636×10⁻⁸ m

Explanation:

Given:

Width of the infinite well, L = 340 pm = 340×10⁻¹² m.

The formula for energy of the electron in nth state is:

$E_n=\frac {n^2\times h^2}{8mL^2}$

The expression for the difference in energy between the levels having quantum numbers n(initial) to n(final) is:

$\Delta E_n=\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}$

According to Planks theory:

E = hv

where, v is the frequency

Also,

Frequency×Wavelength = Speed of light

So,

$E=\frac {hc}{\lambda}$

$\lambda=\frac {hc}{E}$

Also, using energy from above formula as:

$\lambda=\frac {hc}{\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}}$

$\lambda=\frac {c\times {8mL^2}} {({n_f}^2-{n_i}^2)\times h}}$

For longest wavelength ni = 1 and nf = 2

m= mass of the electron = 9.1 ×10⁻³¹kg

c = 3×10⁸m/s

h = 6.625×10⁻³⁴J.sec

$\lambda_{Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({2}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Longest wavelength = 1.2703×10⁻⁷ m

For second longest wavelength ni = 1 and nf = 3

$\lambda_{Second\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({3}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Second longest wavelength = 4.7636×10⁻⁸ m

For third longest wavelength ni = 1 and nf = 4

$\lambda_{Third\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({4}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Third longest wavelength = 2.5406×10⁻⁸ m

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3 years ago

A cog system on the beginning segment of a roller coaster needs to get 29 occupied cars up a 120-m vertical rise over a time int

Rudik [331]

Answer:

Hello your question has some missing part attached below is the missing part

How much work is done by the cog system on the cars?

answer : 2.053 * 10^7 J

Explanation:

Total weight of car ( wT ) = 5900 * 29 = 171100 N

mass of car ( mT ) = wT / g = 171100 / 9.81 = 17459.18 kg

Hence work done by the cog system on the cars

W = ( KE $_{2}$ + PE $_{2}$ ) - ( KE $_{1}$ + PE $_{1}$ )

= ( 1/2 mv^2 + mgh ) - ( 0 )

= ( 1/2 * 17459.18 * 0.50^2) + ( 17459.18 * 9.81 * 120 )

= 2.053 * 10^7 J

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3 years ago

(quick help please??)A mechanic uses a mechanical lift to raise a car. The car weighs 10,200 N. The work required to do this was

GuDViN [60]

I believe It would be 2.2m

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3 years ago