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MAVERICK [17]
3 years ago
15

Exercise 15.1 (a) Find the magnitude of the electric force between two protons separated by 1 femtometer (10^-15), approximately

the distance between two protons in the nucleus of a helium atom. (b) If the protons were not held together by the strong nuclear force, what would be their initial acceleration due to the electric force between them
Physics
1 answer:
Goryan [66]3 years ago
7 0

Answer:

(a) 230.4 N

(b) a = 1.38\times10^{29} m/s^{2}

Explanation:

Charge on a proton, q = 1.6\times10^{-19}C

Distance between the two protons, d = 1 femto meter = 1\times10^{-15}m

(a) By use of coulomb's law, the force between the two protons is given by

F =\frac{k\times q\times q}{d^{2}}

Where, k is the Coulombic constant = 9x 10^9 Nm^2/C^2

So, the force between them is given by

F =\frac{9\times10^{9}\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{\left (1\times10^{-15}  \right )^{2}}

F = 230.4 N

(b) Acceleration is the ratio of force to the mass of proton.

the mass of proton, m = 1.67 x 10^-27 kg

So, acceleration

a=\frac{230.4}{1.67\times10^{-27}}

a = 1.38\times10^{29} m/s^{2}

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miv72 [106K]
B might be the correct answer
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3 years ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

This means that the relation between the wavelength and the length of the string is

3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

4 0
3 years ago
The frequency of a microwave is 1.2 x 10^9 hertz. what is the wavelength of the given problem.
Olenka [21]

Answer:

0.25 m

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field, oscillating in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

c=3.0\cdot 10^8 m/s

Microwave is an example of electromagnetic waves.

The relationship between wavelength and frequency for an electromagnetic wave is:

\lambda=\frac{c}{f}

where

\lambda is the wavelength

c=3.0\cdot 10^8 m/s  is the speed of light

f is the frequency

For the microwave in this problem,

f=1.2\cdot 10^9 Hz

So its wavelength is

\lambda=\frac{3.0\cdot 10^8}{1.2\cdot 10^9}=0.25 m

7 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
Anyone knows this? Please answer... Spam will be reported.
Yakvenalex [24]

Answer:

The correct option is;

The assertion is correct, but reason wrong

Explanation:

The question is with regards to the relationship between work, energy, power, and velocity

The mass of each of the persons running up the staircase = Different

The time it takes each person to run up the stairs = Equal time

Let, 'm₁' and 'm₂' represent the mass of each of the persons that ran up the stairs and m₁ > m₂

Let 't' represent the equal time it takes then to run up the stairs

Let 'h' represent the height of the stairs

The energy, 'E', it takes to run up the stairs is equal to the potential energy, P.E., obtained at the top of the stairs

P.E. = m·g·h

Where;

m = The mass of the person at an elevated height

g = The acceleration due to gravity = Constant

h = The height reached above ground level

Given that the height reached is the same for both of the persons, we have

For m₁, P.E.₁ = m₁·g·h and for m₂, P.E.₂ = m₂·g·h

Therefore, where, m₁ > m₂, we have;

P.E.₁ > P.E.₂

∴ E₁ > E₂

Power, 'P', is the rate at which energy is expended

∴ Power, P = E/t

∴ P₁ = E₁/t  > P₂ = E₂/t

Therefore, the person with the greater mass, 'm₁', uses more power than the person of mass 'm₂', in running up the stairs

Therefore, the assertion is correct

The average velocity, vₐ = (Total distance traveled, d)/(Total time taken, t)

Given that the distance, 'd', covered in running up the stairs by both persons is the same, and the time it takes them to complete the distance, 't', is also the same, we have;

The average velocity of the person with the greater mass m₁ is the same as the average velocity of the person with mass, m₂

Therefore, the reason is wrong

The answer is that the assertion is correct, but reason wrong

6 0
2 years ago
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