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Ira Lisetskai [31]
3 years ago
6

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:
Nana76 [90]3 years ago
7 0

Answer:

E

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice  the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

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How do you do this problem?
mojhsa [17]

Answer:

C

Explanation:

When A and B come in contact with each other, +12 - 12 = 0 so their changes cancel.

Now C has a charge of +12

When A and C come together they each have an equal share of that 12, so each of them has 6

So the answer is

A       B       C

6        0       6

which is C

4 0
3 years ago
A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.
gtnhenbr [62]

Answer:

(a)  vmax = 0.34m/s

(b)  v = 0.13m/s

(c)  v = 0.31m/s

(d)  x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

v_{max}=\omega A       (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

\omega=\sqrt{\frac{k}{m}}           (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

x=Acos(\omega t)

You solve the previous equation for t:

t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s

With this value of t, you can calculate the speed of the object with the following formula:

v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s

The position will be:

x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0
3 years ago
A farmhand attaches a 25-kg bale of hay to one end of a rope passing over a
Georgia [21]

Explanation:

Newton's second law:

∑F = ma

277 N − 245 N = (25 kg) a

a = 1.28 m/s²

8 0
2 years ago
If the electric field of an electromagnetic wave is in the x-direction and the magnetic field of the wave is in the y-direction,
Ulleksa [173]

Answer:

Positive z-direction

Explanation:

According to Fleming's right hand rule, point the fingers of your right hand in the direction of Electric field E ( positive x-direction), and curl your fingers toward magnetic field B (positive y-direction), and your thumb points in the direction of propagation of wave (positive z-direction).

Therefore, the correct option will be positive z-direction.

3 0
3 years ago
What is meant by the term law of conservation?​
GalinKa [24]

Answer:

Explanation:

any law stating that some quantity or property remains constant during and after an interaction or process, as conservation of charge or conservation of linear momentum.

4 0
2 years ago
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