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emmasim [6.3K]
3 years ago
8

Can A positively charged body attract another positively charged body​

Physics
1 answer:
andriy [413]3 years ago
3 0

Like charges repel, unlike charges attract

Two protons will also tend to repel each other because they both have a positive charge. On the other hand, electrons and protons will be attracted to each other because of their unlike charges.

So I would say no, unless the two bodies are placed close to each other where one has much more charge than the other, then due to induction, force of attraction becomes more than the force of repulsion.

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Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o
Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.  

\lambda_{proton} = 6.05x10^{-14}m < \lambda_{electron} = 1.10x10^{-10}m

Part B:

The proton has a smaller wavelength than the electron.

\lambda_{proton} = 1.29x10^{-13}m < \lambda_{electron} = 5.525x10^{-12}m

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

\lambda = \frac{h}{p} (1)

Where h is the Planck's constant and p is the momentum.

\lambda = \frac{h}{mv} (2)

Part A

Case for the electron:

\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

But J = Kg.m^{2}/s^{2}

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}

\lambda = 1.10x10^{-10}m

Case for the proton:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}

\lambda = 6.05x10^{-14}m

Hence, the proton has a smaller wavelength than the electron.  

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

KE = \frac{1}{2}mv^{2}

2KE = mv^{2}

v^{2} = \frac{2KE}{m}

v = \sqrt{\frac{2KE}{m}}  (3)

Case for the electron:

v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}

but 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}

v = 1.316x10^{8}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}    

\lambda = 5.525x10^{-12}m

Case for the proton :

v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}

But 1J = kg \cdot m^{2}/s^{2}

v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}

v = 3.07x10^{6}m/s

Then, equation 2 can be used:

\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}

\lambda = 1.29x10^{-13}m    

Hence, the proton has a smaller wavelength than the electron.

7 0
3 years ago
620 J of heat is added to the cylinder of an engine, which causes the gas inside to expand. As a result, the piston of the engin
Maru [420]

Answer:

<u>400</u> J work is done BY the engine.

The internal energy of the gas is <u>620</u> J

Explanation:

The given information are;

The heat added to the cylinder = 620 J

The force applied by the piston of the engine = 8.0 kN = 8,000 N

The distance over which the force moves (the piston) = 5.0 cm = 0.05 m

The work done (by the engine) = Force × Distance = 8,000 N × 0.05 m = 400 J

The internal energy is the sum of the kinetic and potential energy of the system

Therefore, given that the internal energy, U, is the sum total of the energy in the system

∴ U = The heat supplied to the system = 620 J

Which gives;

<u>400</u> J work is done BY the engine.

The internal energy of the gas is <u>620</u> J.

6 0
2 years ago
A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its negative terminal. How much en
dem82 [27]

Answer:

E = 0.18 J

Explanation:

given,

Potential of the battery,V = 9 V

Charge on the circuit, Q = 20 m C

                                        = 20 x 10⁻³ C

energy delivered in the circuit

E = Q V

E = 20 x 10⁻³ x 9

E =  180 x 10⁻³

E = 0.18 J

Energy delivered in the circuit is equal to E = 0.18 J

7 0
3 years ago
It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo
vovikov84 [41]

Answer:

L = 1.11 x 10^{6} m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x 10^{5} m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x 10^{-2} m

Length = ?

So,

we know that,

U = 1/2 C Δv^{2}

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δv^{2}  

52000 J = (0.5) x (C) x (610^{2})

C = 0.28 F

And we also know that,

C = \frac{K*E*A}{d}

E = 8.85 x 10 ^{-12}

K = 3.7

A = 0.20 x L

d = 2.6 x 10^{5} m

Plugging in the values into the formula, we get:

0.28 = \frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }

Solving for L, we get:

L = 1.11 x 10^{6} m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.  

7 0
3 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
3 years ago
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