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3 years ago

11

What is the electric flux Φ3 through the annular ring, surface 3? Express your answer in terms of C , r1, r2, and any constants.

1 answer:

mr Goodwill [35]3 years ago

4 0

Answer:

Pls see attached file

Explanation:

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A gamma-ray photon produces an electron and a positron, each with a kinetic energy of 261 keV . h=6.626×10−34J⋅s, c=2.998×108m/s

ddd [48]

Answer:

The energy and the wavelength of the photon are 1.546 MeV and $8.036\times10^{-13}\ m$ .

Explanation:

Given that,

Kinetic energy = 261 KeV

Planck's constant $h = 6.626\times10^{−34}\ J.s$

Speed of light $c=2.998\times10^{8}\ m/s$

Mass of electron $m_{e}=9.109\times10^{-31}\ kg$

Charge $q=1.602\times10^{-19}\ C$

(A). We need to calculate the energy of the photon

Using formula of rest mass energy

$E=m_{0}c^2$

$E=9.109\times10^{-31}\times(3\times10^{8})^2$

$E=8.198\times10^{-14}\ J$

Energy in eV

$E=\dfrac{8.198\times10^{-14}}{1.6\times10^{-19}}$

$E=512375\ eV$

$E=0.512\ MeV$

The total energy of photon

$TE=2(E+K.E)$

$TE=2(0.512+0.261)$

$TE=1.546\ MeV$

(B). We need to calculate the wavelength of the photon

Using formula of wavelength

$\lambda=\dfrac{hc}{E}$

Put the value into the formula

$\lambda=\dfrac{6.626\times10^{−34}\times3\times10^{8}}{1.546\times10^{6}\times1.6\times10^{-19}}$

$\lambda=8.036\times10^{-13}\ m$

Hence, The energy and the wavelength of the photon are 1.546 MeV and $8.036\times10^{-13}\ m$ .

7 0

3 years ago

A CAR APPROACHING A SCHOOL ZONE SLOWS DOWN FROM 38 M/S TO 10 M/S AT A CONSTANT ACCELERATION -4 METER PER SEC SQUARE HOW MUCH TIM

Andre45 [30]

Answer:

7 seconds

Explanation:

he went down -28 m/s so 7 times -4 equals -28 m/s which is what he needed to slowness down to 10 m/s

8 0

3 years ago

a car accelerates uniformly at from rest at a speed of 23.7 km/h in 6.5s. find the distance the car travels during this time

hodyreva [135]

speed = distance / time

distance =speed × time

d = v × t

d= 23.7 kmh⁻¹ ˣ 6.5 s

d = 154.05 kmh⁻¹s.

3 0

4 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

Travka [436]

Yes you are correct...I dont think you need to do this? You are getting them all right but now your just wasting points..

But anyways you are correct. ♡♡

5 0

4 years ago

A comet with a mass of 7.85 × 1011 kg is moving with a velocity of 25,000 m/s. calculate its kinetic energy.

Nadusha1986 [10]

The kinetic energy of an object is given by:
$K= \frac{1}{2}mv^2$
where
K is the kinetic energy
m is the mass of the object
v is its velocity.

The comet in our problem has a mass of $m=7.85 \cdot 10^{11} kg$ and a velocity of $v=25000 m/s$ , so its kinetic energy is:
$K= \frac{1}{2}(7.85 \cdot 10^{11} kg)(25000 m/s)^2=2.45 \cdot 10^{20}J$

6 0

3 years ago