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DerKrebs [107]
4 years ago
9

What is the wavelength of a proton traveling at a speed of 6.21 km/s? what would be the region of the spectrum for electromagnet

ic radiation of this wavelength?
Physics
1 answer:
lukranit [14]4 years ago
8 0
<span>The wavelength of a proton traveling at a speed of 6.21 km/s is wavelength = (6.63E-34 Js) / (1.67Eâ’27kg x 6.21E3m/s) = 6.393E-9 m Region of the spectrum for electromagnetic radiation of this wavelength will be Xrays.</span>
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Tick (3) the correct statement about electrostatic charges.
Radda [10]

Answer:

similar type of electric charges attract one another

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3 years ago
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Cierto volumen de gas se encuentra a 60°c de temperatura y 5atm de presión, es calentado hasta 140°c, estado en el cual ocupa un
earnstyle [38]

Answer:

V1 = 2221.33 L

Explanation:

The system is about a ideal gas. Then you can use the equation for ideal gases for a volume V1, temperature T1 and pressure P1:

P_1V_1=nRT_1   (1)

And also for the situation in which the variables T, V and P has changed:

P_2V_2=nRT_2   (1)

R: constant of ideal gases = 0.082 L.atm/mol.K

For both cases (1) and (2) the number of moles are the same. Next, you solve for n in (1) and (2):

n=\frac{P_1V_1}{RT_1}\\\\n=\frac{P_2V_2}{RT_2}

Next, you equal these equations an solve for T2:

\frac{P_1V_1}{RT_1}=\frac{P_2V_2}{RT_2}\\\\V_1=\frac{P_2V_2T_1}{P_1T_2}

Finally you replace the values of P2, V2, T1 and T2:

V_1=\frac{(7atm)(680L)(140\°C)}{(60\°C)(5atm)}=2221.33\ L

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5 0
4 years ago
A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is. (a) How m
Gelneren [198K]

Answer:

a. 5 × 10¹⁹ protons b. 2.05 × 10⁷ °C

Explanation:

Here is the complete question

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.42 A. (a) How many protons strike the target in 19 seconds? (b) Each proton has a kinetic energy of 6.0 x 10-12 J. Suppose the target is a 17-gram block of metal whose specific heat capacity is 860 J/(kg Co), and all the kinetic energy of the protons goes into heating it up. What is the change in temperature of the block at the end of 19 s?

Solution

a.

i = Q/t = ne/t

n = it/e where i = current = 0.42 A, n = number of protons, e = proton charge = 1.602 × 10⁻¹⁹ C and t = time = 19 s

So n = 0.42 A × 19 s/1.602 × 10⁻¹⁹ C

       = 4.98 × 10¹⁹ protons

       ≅ 5 × 10¹⁹ protons

b

The total kinetic energy of the protons = heat change of target

total kinetic energy of the protons = n × kinetic energy per proton

                                                         = 5 × 10¹⁹ protons × 6.0 × 10⁻¹² J per proton

                                                         = 30 × 10⁷ J

heat change of target = Q = mcΔT ⇒ ΔT = Q/mc where m = mass of block = 17 g = 0.017 kg and c = specific heat capacity = 860 J/(kg °C)

ΔT = Q/mc = 30 × 10⁷ J/0.017 kg × 860 J/(kg °C)

     = 30 × 10⁷/14.62

     = 2.05 × 10⁷ °C

5 0
3 years ago
Which of the following is an example of deduction
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pogonyaev
T = 3.5 secs

Velocity (v) = g * t = 10 m/s^2 * 3.5 sec = 35 m/s
7 0
3 years ago
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