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Ronch [10]
2 years ago
12

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 47000 m/s. The experiment is repeated with a He ion (charge e, mass 4 u).
What is the ion's speed at the negative plate?
Physics
1 answer:
WINSTONCH [101]2 years ago
7 0

Answer:

=2,012,319.36 \ m/s

Explanation:

-The only relevant force is the electrostatic force

-The formula for the electrostatic force is:

F = Eq

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

#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 magnitude of the electric forces acting in both proton and electron are the same.

F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton

-Applying Newton's 2nd Law:

F=ma

F_e=M_ea_e

F_p=M_pa_p

#equate the two forces:

F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}

#The equations for velocity in uniform acceleration:

V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad

#For the proton:

V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}

#For the electron:

V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}

The mass values of the proton and electron are:

M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg

The speed of the ion is therefore calculated as:

V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s

Hence, the ion's speed at the negative plate is =2,012,319.36 \ m/s

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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
kobusy [5.1K]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere  ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon  }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Q₁ = 572.8 * 10^{-9} C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2}  = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  =0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
2 years ago
A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put
olga_2 [115]

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

F_s = F_g

kx = mg

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

k = \frac{mg}{x}

k = \frac{8*9.8}{0.1}

k = 784N/m

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

PE = \frac{1}{2} kx^2

PE =\frac{1}{2} 784*0.4^2

PE = 63.72J

Therefore the energy stored in the spring is 63.72J

6 0
2 years ago
A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N. If instead the sample
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Answer:128 N

Explanation:

Sample of 3 cm and 4 mm diameter found to break under a minimum force of 128 N .

If sample is 1.5 cm long with same cross-sectional area then minimum force required to break is also 128 N because the applied force is same for any length and diameter of tendon.        

3 0
2 years ago
Discuss the production, transmission, and usage of electricity in the context of conservation of energy. When electricity is “us
mr Goodwill [35]

There are huge losses in the transmission, production and usage of electricity and the reduction of these losses in order to save electricity is called as conservation of energy.

As per the statistics, there is loss of nearly 4% while the transmission of electricity. Like wise during production also, lot of electricity get wasted due to the inefficient material used. None of the production material nor the equipment used have 100% efficiency and thus there is always a possibility of energy wastage.

When it is said that the energy is wasted , it simply means that the energy production which should have been 100% as per calculation is not completely derived from the source due to the inefficient conversion process. For example, a turbine while rotating must convert 100 % of the water energy or water falling on it into electrical energy but the turbine is not able to do so as some of the water is lost or its energy is lost before conversion while going through the mechanical process.

3 0
2 years ago
A heavy frog and a light frog jump straight up into the air. They push off in such away that they both have the same kinetic ene
Ilia_Sergeevich [38]

Answer:

The lighter frog goes higher than the heavier frog.

The lighter frog is moving faster than the heavier frog

Explanation:

If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

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vf^{2} -vo^{2} = 2*a*hmax = vf^{2} -vo^{2} = 2*(-g)*hmax

When h= hmax, the object comes momentarily to an stop, so vf =0

Solving for hmax:

hmax =\frac{vo^{2} }{2*g}

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

5 0
2 years ago
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