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zimovet [89]
3 years ago
13

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

ative plate with a speed of 55000m/s . What will be the final speed of an electron released from rest at the negative plate?
Physics
2 answers:
Natali5045456 [20]3 years ago
5 0

Answer: Vf = 2,400,000 m/s

Explanation:

1) The only relevant force is the electrostatic force

2) The formula for the electrostatic force is F = E×q

Where E is the electric field and q is the magnitude of the charge.

3) Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magntitude of the electric forces acting in both proton and electron are the same

4) Fe = Fp (Fe stands for force on the electron and Fp stands for force on the proton).

5) Using second law of Newton, Force = mass × acceleration

Fe = Me × Ae (Me is mass of the electron, and Ae is acceleration of the electron)

Fp = Mp × Ap (Mp is mass of the proton, and Ap is acceleration of the proton)

⇒Me × Ae = Mp × Ap

⇒ Ae = Mp × Ap / Me

6) Now, state the equations for the velocity in uniformly accelerated motion:

i) Vf² = Vo² + 2ad

Vo² = 0 for both cases, and d is the same distance.

⇒ Vf² = 2ad

ii) For the proton Vf² = 2(Ap)(d) ⇒ Ap = Vf² / (2d)

⇒ Ap = (55,000 m/s)² / (2d)

iii) For the electron Vf² = 2(Ae)² (2d)

iv) Using Ae = Mp × Ap / Me (found prevously):

Vf² = Mp × (55,000 m/s)² / (2d) × (2d) / Me

⇒ Vf² = Mp × (55,000 m/s)² / Me

Taking square root in both sides:

⇒ Vf = 55,000 m/s × √ [Mp / Me]

7) These are the values for the masses of a proton and an electron:

Mp = 1.67 × 10⁻²⁷ kg

Me = 9.11×10⁻³¹ kg

8) Replace and compute:

Vf = 55,000 m/s × √ [ 1.67 × 10⁻²⁷ kg / 9.11×10⁻³¹ kg] = 2,354,841.8 m/s

Round to two significan digits: Vf = 2,400,000 m/s

SVEN [57.7K]3 years ago
3 0
The correct answer to this question is this one:

You will need these relations

<span>qV = KE = 1/2m<span>v^2

</span></span>

You know the charge of a proton (the elementary charge in this case), you know its mass, and you know it's velocity. From there you can find its kinetic energy and from there you can find it's voltage. Once you know the potential difference there all you have to do is reverse the process. Multiply the charge of an electron by V and then solve for the velocity using the mass of the electron

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Please helps me with this question..
Alenkasestr [34]
1) S.I. Unit for electric current = "Ampere"

2) S.I. Unit for resistance = "Ohm"

3) S.I. Unit for potential difference = "Volt"

Hope this helps!
8 0
3 years ago
Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang
Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

                                         ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

                              m*g*s*sin(θ) = 0.25*m*R^2*w^2

                              2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0
3 years ago
What changes must be done to the wire to increase its conductance.​
777dan777 [17]

Answer:

- Decreasing the resistance

- Using a shorter length

- Using a smaller area wire

Explanation:

Formula for conductance in wires is;

G = 1/R

Where;

G is conductance

R is resistance

This means that increasing the resistance leads to a larger denominator and thus a smaller conductance but to decrease the denominator means larger conductance.

Thus, to increase the conductance, we have to decrease the resistance.

Resistance here has a formula of;

R = ρL/A

Where;

ρ is resistivity

L is length of wire

A is area

Thus, to decrease the resistance, we will have to use a shorter length and smaller area of wire.

8 0
3 years ago
A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.
ANTONII [103]

<u>Answer:</u> The number of photons are 3.7\times 10^8

<u>Explanation:</u>

We are given:

Wavelength of microwave = 1.22\times 10^8nm=0.122m    (Conversion factor:  1m=10^9nm  )

  • To calculate the energy of one photon, we use Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.625\times 10^{-34}J.s

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 0.122 m

Putting values in above equation, we get:

E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J

  • To calculate the mass of water, we use the equation:

Density=\frac{Mass}{Volume}

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g

  • To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

q=mc\Delta T

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = T_2-T_1=100^oC-23^oC=77^oC

Putting values in above equation, we get:

q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J

This energy is the amount of energy for 'n' number of photons.

  • To calculate the number of photons, we divide the total energy by energy of one photon, we get:

n=\frac{q}{E}

q = 53127.72 J

E = 1.44\times 10^{-24}J

Putting values in above equation, we get:

n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}

Hence, the number of photons are 3.7\times 10^8

4 0
3 years ago
. As we increase the quantum number of an electron in a one-dimensional, infinite potential well, what happens to the number of
Natalija [7]

Answer:

It increases.

Explanation:

For the electron to escape the photon needs energy is equal to the difference between initial and its non quantised region energy , then only it will be able to escape finite well.

E ∝ n^2

n= energy state quantum number

so if , n increases maximum point of probability density increases.

8 0
3 years ago
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