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Georgia [21]
3 years ago
5

When light passes into a more dense material, it bends away from the

Physics
2 answers:
asambeis [7]3 years ago
7 0

Answer: true

Explanation:

Ne4ueva [31]3 years ago
5 0

Answer:

its true (~ ̄▽ ̄)~

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What is meant by the phrase " a consistent method of measurement "? someone help plss
Mars2501 [29]
<span><span>D.</span><span>Measurements are taken in a way that is the same every time.- apex

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8 0
3 years ago
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Who speaks the line "Lord, what fools these mortals be"?
ivanzaharov [21]
The answer is D.Puck.

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3 0
3 years ago
An electron is confined to a one dimensional region, bounded by an infinite potential. If the energy of the electron in its firs
OLga [1]

Answer:

The energy in its ground state is 10 meV.

Explanation:

It is given that,

The energy of the electron in its first excited state is 40 meV.

Energy of the electron in any state is given by :

E=\dfrac{n^2\pi^2h^2}{8mL^2}

For ground state, n = 1

E_1=\dfrac{\pi^2h^2}{8mL^2}.............(1)

For first excited state, n = 2

40=\dfrac{2^2\pi^2h^2}{8mL^2}.............(2)

Dividing equation (1) and (2), we get :

\dfrac{E_1}{40}=\dfrac{1}{4}

E_1=10\ meV

So, the energy in its ground state is 10 meV. Hence, this is the required solution.

4 0
3 years ago
Ammonia gas occupies a volume of 0.450 L at a pressure of 96 kPa. What
Sauron [17]

Answer:

0.426 L

Explanation:

Boyles law is expressed as p1v1=p2v2 where

P1 is first pressure, v1 is first volume

P2 is second pressure, v2 is second volume.

Given information

P1=96 kPa, v1=0.45 l

P2=101.3 kpa

Unknown is v2

Making v2 the subject from Boyle's law

v2=\frac {p1v1}{p2}

Substituting the given values then

v2=\frac {96*0.45}{101.3}=0.4264560710760l\approx 0.426 l

Therefore, the volume is approximately 0.426 L

6 0
2 years ago
510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the dr
Yuki888 [10]

Answer:

The terminal velocity is v_t  =17.5 \ m/s

Explanation:

From the question we are told that

       The mass of the squirrel is  m_s  =  50\ g  =  \frac{50}{1000} =  0.05 \  kg

      The surface area is   A_s =  935 cm^2  =  \frac{935}{10000} = 0.0935 \ m^2

       The height of fall is  h =4.8 m

        The length of the prism is l =  23.2 = 0.232 \ m

          The width of the prism is w =  11.6 =  0.116 \ m

 

The terminal velocity is mathematically represented as

       v_t  =  \sqrt{\frac{2 * m_s *  g }{\dho_s * C  * A } }

Where \rho  is the density of a rectangular prism with a constant values of \rho  =  1.21 \ kg/m^3

            C is the drag coefficient for a horizontal skydiver with a value = 1

            A  is the area of the prism the squirrel is assumed to be which is mathematically represented as

      A =  0.116 * 0.232

       A =  0.026912 \ m^2

 substituting values

      v_t  =  \sqrt{\frac{2 * 0.510 *  9.8 }{1.21 * 1  * 0.026912 } }

     v_t  =17.5 \ m/s

       

7 0
3 years ago
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