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3 years ago

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A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B

roglie wavelength λ at the end of this period?

1 answer:

Novay_Z [31]3 years ago

4 0

Answer:

Wavelength, $\lambda=1.28\times 10^{-14}\ m$

Explanation:

Given that,

Mass of the particle, $m=4.3\times 10^{-28}\ kg$

Acceleration of the particle, $a=2.4\times 10^7\ m/s^2$

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

$\lambda=\dfrac{h}{mv}$

v = a × t

$\lambda=\dfrac{h}{mat}$

$\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}$

$\lambda=1.28\times 10^{-14}\ m$

Hence, this is the required solution.

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Setler [38]

I think the answer is B.

Hope this helps.

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3 years ago

Read 2 more answers

A car has an acceleration of -5 m/s^2. Describe the car’s motion

AlladinOne [14]

Explanation:

because the acceleration is negative, this indicates a deceleration (or slowing down) . Hence we can say that:

The car is decelerating (slowing down), i.e its velocity is decreasing, at a constant rate of 5m/s².

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3 years ago

Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr

lesya692 [45]

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time of computer A is $CT_{A} =200 ps$

Clock time of computer B is $CT_{B} =250 ps$

Effective CPI of computer A is $CPI_{A} =1.5$

Effective CPI of computer B is $CPI_{B} =1.7$

CPU time of A is

$CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec$

CPU time of B is

$CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec$

Hence Computer A is Faster by $\frac{425}{300} =1.41$

Computer A is 1.41 times faster than the Computer B

4 0

3 years ago

The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the

vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r² where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

I₂ = 2.1 x 10⁹ W/m²

4 0

3 years ago

A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle

Contact [7]

<h2>Answers: </h2>

1) 1.359, 1.403

2) $2.207(10)^{8}m/s$ , $2.138(10)^{8}m/s$

Explanation:

The described situation is known as Refraction.

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index $n$ is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ( $c=3(10)^{8}m/s$ ) and the speed of light $v$ in the second medium:

$n=\frac{c}{v}$ (1)

On the other hand we have the Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (2)

Where:

$n_{1}$ is the first medium refractive index . We are told is the air, hence $n_{1}\approx 1$

$n_{2}$ is the second medium refractive index

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=38.1\º$ :

$(1)sin(57.0\º)=n_{2}sin(38.1\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}$

$n_{2}=1.359$ (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=36.7\º$ :

$(1)sin(57.0\º)=n_{2}sin(36.7\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}$

$n_{2}=1.403$ (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

$n_{red}=\frac{c}{v_{red}}$

$v_{red}=\frac{c}{n_{red}}$

Substituting (3) in this equation:

$v_{red}=\frac{3(10)^{8}m/s}{1.359}$

$v_{red}=2.207(10)^{8}m/s$ >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

$n_{violet}=\frac{c}{v_{violet}}$

$v_{violet}=\frac{c}{n_{violet}}$

Substituting (4) in this equation:

$v_{violet}=\frac{3(10)^{8}m/s}{1.403}$

$v_{red}=2.138(10)^{8}m/s$ >>>>Speed of violet light

3 0

3 years ago