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4 years ago

13

A gamma ray with an energy of 3.40 × 10-14 joules strikes a photographic plate. We know that Planck's constant is 6.63 × 10-34 j

oule·seconds. What is the frequency of the photon?
5.13 × 1019 hertz

5.60 × 1017 hertz

2.45 × 1017 hertz

2.45 × 10-12 hertz

7.80 × 10-19 hertz

2 answers:

suter [353]4 years ago

4 0

The correct answer is: "A. 5.13 x 10^19 hertz"

Wittaler [7]4 years ago

4 0

The answer is A, 5.13 x 10^19 hertz

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You drive 200 miles in 3 hours before stopping for lunch and gas. After lunch you

Paraphin [41]

Given :

You drive 200 miles in 3 hours before stopping for lunch and gas. After lunch you travel 250 miles in an hour and a half.

To Find :

Average speed.

Solution :

We know, average speed is given by :

$v=\dfrac{\text{Total distance covered}}{\text{Total time taken}}\\\\v=\dfrac{200+250}{3+1.5}\ miles/hr\\\\v=\dfrac{450}{4.5}\ miles/hr\\\\v=100\ miles/hr$

Therefore, average speed of the journey is 100 miles/hr.

Hence, this is the required solution.

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4 years ago

What properties of sound determine the volume of sound? Is this affected by the motion of the sound source?

TEA [102]

The property of sound that determine its volume is its AMPLITUDE. The amplitude and the volume of the sound are directly varied. Which means that the larger the amplitude, the larger is the volume and the softer it is if the amplitude is smaller. It is not affected by the motion of the sound source so long as the barriers are not breached.

8 0

3 years ago

Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the

Vinil7 [7]

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

The energy equation is:

$q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +delta-p$

For a diffuser, w = Δp = 0

The diffuser is adiabatic, q = 0

Replacing:

$0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +0$

Where

V₁ = 250 m/s

V₂ = 40 m/s

Replacing:

$0-0=h_{2} -400x10^{3} +\frac{40^{2}-250^{2} }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg$

Using tables, at 431.43 kJ/kg the temperature is 430 K

b) The inlet area is:

$m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}$

The exit area is:

$m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}$

6 0

3 years ago

The motor on a helicopter turns at an angular speed of 6.2 x 102 revolutions per minute. (a) Express this angular speed in radia

arsen [322]

Answer:

64.93 rad/s
38.956 km

Explanation:

(a)

$\dfrac{6.2\cdot 10^2\,\text{rev}}{60\,\text{s}}\times\dfrac{2\pi\,\text{rad}}{\text{rev}}=\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\approx\boxed{64.93\,\text{rad/s}}$

__

(b)

$d=r\theta=(3.0\,\text{m})(2.0\cdot 10^2\,\text{s})\left(\dfrac{62\pi\,\text{rad}}{3\,\text{s}}\right)=124\pi\cdot 10^2\,\text{m}=\boxed{38\,956\,\text{m}}$

6 0

3 years ago

A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV.

Mrrafil [7]

Answer:

The magnetic field required required for the beam not to be deflected is $B = 0.0036T$

Explanation:

From the question we are told that

The charge on the particle is $q = +2e$

The mass of the particle is $m = 6.64 *10^{-27} kg$

The potential difference is $V_a = 1.8 kV = 1.8 *10^{3} V$

The potential difference between the two parallel plate is $V_b = 120 V$

The separation between the plate is $d = 8 mm = \frac{8}{1000} = 8*10^{-3}m$

The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam after the region having a potential difference of 1.8kV

$KE_b = PE_b$

Generelly

$KE_b = \frac{1}{2} m v^2$

And $PE_b = q V_a$

Equating this two formulas

$\frac{1}{2} mv^2 = q V_a$

making v the subject

$v = \sqrt{\frac{q V_a}{2 m} }$

Substituting value

$v = \sqrt{\frac{ 2* 1.602 *10^{-19} * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }$

$v = 41.65*10^4 m/s$

Generally the electric field between the plates is mathematically represented as

$E = \frac{V_b}{d}$

Substituting value

$E = \frac{120}{8*10^{-3}}$

$E = 15 *10^3 NC^{-1}$

the magnetic field is mathematically evaluate

$B = \frac{E}{v}$

$B = \frac{15 *10^{3}}{41.65 *10^4}$

$B = 0.0036T$

6 0

3 years ago