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REY [17]
3 years ago
14

The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

ectric field at twice the distance from the point charge?
a.) At twice the distance, the strength of the field is E/2.
b.) At twice the distance, the strength of the field is 2E.
c.)At twice the distance, the strength of the field is 4E.
d.)At twice the distance, the strength of the field remains equal to E.
e.)At twice the distance, the strength of the field is E/4.
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
4 0

Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}

so, at twice the distance the strength of the field is E/4.

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Flammable materials, like alcohol, should never be dispensed or used near
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A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is
BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

<em>In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  </em>

<em>In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  </em>

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

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<u>Ques</u><u>tion</u><u>:</u>

Azaria's pet lizard runs a distance of 45 meters in 24 seconds. What is the speed of the lizard?

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<u>Solu</u><u>tion</u><u>:</u>

We are given,

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Speed is given by:

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_________________________________________________

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