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Viefleur [7K]
3 years ago
8

The particle accelerator at CERN can accelerate an electron through a potential

Physics
1 answer:
asambeis [7]3 years ago
8 0

Answer:

K.E = 1.28 × 10^-17 KeV

Explanation:

Given that a particle accelerator at CERN can accelerate an electron through a potentialdifference of 80 kilovolts.

To Calculate the kinetic energy (in keV) of the electron​, let us first find the electron charge which is 1.60 × 10^-19C

The kinetic energy = work done

K.E = e × kV

Substitute e and the voltage into the formula

K.E = 1.60 × 10^-19 × 80

K.E = 1.28 × 10^-17 KeV

Therefore, the kinetic energy is approximately equal to 1.28 × 10^-17 KeV

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What effect or effects would be most significant if the Moon's orbital plane were exactly the same as the ecliptic plane?
Lilit [14]

Answer:

c. Solar eclipses would be much more frequent.

Explanation:

The <u>ecliptic plane</u> is the apparent orbit that the sun describes around the earth (although it is the earth that orbits the sun), is the path the sun follows in earth's sky.

A <u>solar eclipse</u> occurs when the moon gets between the earth and the sun, so a shadow is cast on the earth because the light from the sun is blocked.

The reason why solar eclipses are not very frequent is because the moon's orbital plane is not in the same plane as the orbit of the earth around the sun, but rather that it is somewhat inclined with respect to it.

So <u>if both orbits were aligned, the moon would interpose between the sun and the earth more frequently, producing more solar eclipses.</u>

So, if the moon's orbital plane were exacly the same as the ecliptic plane solar eclipses would be more frequent.

the answer is: c.

8 0
4 years ago
Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem
statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

x=x_o+v_1t    (1)

xo = 10 km

v1 = 55km/h

car 2:

x'=v_2t    (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}

t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s

The position in which both cars coincides is:

x=(55km/h)(0.33h)=18.33km

6 0
3 years ago
15. You are watching a baseball game on television that 15
Gre4nikov [31]

Answer:

t_total = 6.99 s

Explanation:

It asks us how long it takes to hear the sound, for this we must look for the time (t₁) it takes for the sound to reach the microphone, the time it takes for the video signal (t₂) to reach the television and the time (₃) it takes for the TV sound to reach us, so the total delay time is

         t_total = t₁ + t₂ + t₂

we look for t1, it indicates that the distance x = 22m

          v = x / t

           t = x / v

           t₁ = 22/343

           t₁ = 6.41 10-2 s

time t₂

          t₂ = 4500 103/3 108

          t₂ = 1.5 10-5 s

time t₃

          t₃ = 2/343

          t₃ = 5.83 10⁻³

Total time is

         t_total = t₁ + t₂ + t₃

         t_total = 6.41 10⁻² + ​​1.5 10⁻⁵ + 0.583 10⁻²

         t_total = 6.99 s

5 0
3 years ago
A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6
Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6  = V/0.0001

V = 2.0 × 10^6  * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q =  1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0
3 years ago
Questions
astra-53 [7]

Answer:

1968

Explanation:

2400*20.5*0.004

8 0
3 years ago
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