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Strike441 [17]
2 years ago
3

Our text claims that a charged particle exerts a net attractive force on an electric dipole. The purpose of this exercise is to

investigate this phenomenon. Suppose a permanent dipole is situated at the origin of the r-z plane and consists of charges +???? and −???? separated by a fixed distance ????. The dipole is situated such that +???? lies at − ???? 2 on the r-axis and −???? lies at ???? 2 on the r-axis. A charge +???? is situated at distance r from the center of the dipole on the positive raxis. We’ll assume, as is usually the case for electric dipoles, that ???? ≪ ???? for this configuration. a. (1 point) Draw a pictorial representation of this physical situation and label any quantities of interest. b. (2 points) Write an expression for the net force exerted on the dipole by charge +????. c. (1 point) Is this force toward +???? or away from +????? Explain. d. (3 points) Use the binomial expansion as an approximation tool to understand the effect that ???? ≪ ???? has on the physical situation. The binomial approximation to first order is wr

Physics
1 answer:
Nostrana [21]2 years ago
7 0

Answer:

Explanation:

Find attached, the solution

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Light from a lamp is shining on a surface. how can you increase the intensity of the light on the surface?
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The intensity of a light in a surface follows the inverse square law formula which can be mathematically expressed as,
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where I is intensity, d is distance, and k is the proportionality constant. For us to increase the intensity, we should lower the distance from the source to the surface. 
4 0
2 years ago
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0
2 years ago
A target in a shooting gallery consists of a vertical square wooden board, 0.250 m on a side and with mass 0.750 kg, that pivots
Alenkasestr [34]

Here in this case since there is no torque about the hinge axis for the system of bullet and block then we can say that angular momentum of this system will remain conserved

L_i = L_f

mv \frac{L}{2} = (I_1 + I_2)\omega

here we will have

L = 0.250 m

v = 385 m/s

m = 1.90 gram

now moment of inertia of the plate will be

I_1 = \frac{ML^2}{3}

I_1 = \frac{0.750 (0.250)^2}{3} = 0.0156 kg m^2

I_2 = m(\frac{L}{2})^2 = 0.0019(0.125)^2 = 2.97 \times 10^{-5} kg m^2

now from above equation

0.0019 (385)(0.125) = (0.0156 + 2.97 \times 10^{-5})\omega

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8 0
2 years ago
A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to
daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s                                                                                

Explanation:

It is given that,

Mass of cart 1, m_1=2\ kg

Mass of cart 2, m_2=4\ kg  

Initial speed of cart 1, u_1=3\ m/s          

Initial speed of cart 2, u_2=0 (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}        

V=\dfrac{2\times 3}{(2+4)}

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

J=F\times t=m(V-u)

J=m(V-u)

J=2\times (1-3)    

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0
2 years ago
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irina1246 [14]
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6 0
2 years ago
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