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maria [59]
2 years ago
7

8. Find the momentum of a photon in eV/c and in Kg. m/s if the wavelength is (a) 400nm ; (b) 1 Å = 0.1 nm, (c) 3 cm ; and (d) 2

nm .
Physics
1 answer:
nataly862011 [7]2 years ago
7 0
We use the formula: p = E/c where E = hc / λ. hence, p = h/  λ. where h is the Planck's constant: 6.62607004 × 10-34 m2 kg / s and <span>λ is the wavelenght. 
</span>
a) p = <span>6.62607004 × 10-34 m2 kg / s / 0.1 x10^-9 m = 6.62607 x 10-24 m kg/s
</span>b) p = 6.62607004 × 10-34 m2 kg / s / 3 x10^-2 m = 2.20869 <span>x 10-32 m kg/s
</span>b) p = 6.62607004 × 10-34 m2 kg / s / 2 x10^-9 m = 3.3130 <span>x 10-25 m kg/s</span>
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A transmission channel is made up of three sections. The first section introduces a loss of 16dB, the second an amplification (o
AlekseyPX

Answer:

P_{out} = 0.100 W = 100 mW

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB

so, the input signal could be expressed in dB terms

P_{in} [dB] = 10 log_{10}(P_{in}) \\P_{in} [dB] = 10 log_{10}(0.4)\\P_{in} [dB] = -3.97 dB (1)

so the output signal could be expressed as.

P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB

The gain should be expressed in dB terms and power in dBm terms so

P_{out} = -9.97 + 30 = 20.03 dBm

using the (1) equation to find it in terms of Watts

P_{out} = 0.100 W = 100 mW

3 0
2 years ago
Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot
Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                  m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0

                  (m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0

Taking out ‘mg’ as common and we get

                  2 x-50=0

                  2 x=50

                  x=\frac{50}{2}=25 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

                 x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                   4 m g(x)+(m g)(x-50 c m)=0

                   4 m g x+m g x-50 m g=0

Taking out ‘mg’ as common and we get

                   5 x=50

                   x=\frac{50}{5}=10 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

                   x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0
2 years ago
A 145-g baseball is thrown so that it acquires a speed of 25 m/s. What was the net work done on the ball to make it reach this s
inysia [295]

When the ball has left your hand and is flying on its own, its kinetic energy is

KE = (1/2) (mass) (speed²)

KE = (1/2) (0.145 kg) (25 m/s)²

KE = (0.0725 kg) (625 m²/s²)

<em>KE = 45.3 Joules</em>

If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from:  It MUST have come from the hand that threw the ball.  The hand would have needed to do  <em>45.3 J</em>  of work on the ball before releasing it.

6 0
2 years ago
A CCD has a greatest possible pixel value of 4095. what is the bit level of this CCD?
Shtirlitz [24]
<span>Bit level for a CCD (Charged coupled device) with a greatest possible pixel value of 4095:The relationship between the bit level and pixel value is given as:pixel value = 2^bit level.Most charged coupled devices (CCDs) have 8-bit, 16-bit, 32-bit levels.Using simple mathematics, we can see that 2^12 = 4096.Since the maximum number of pixels is 4095, the bit level is 12., i.e. the CCD has 12-bit level.</span>
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2 years ago
two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire
valentinak56 [21]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>
6 0
2 years ago
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