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Elodia [21]
2 years ago
9

A rod 12.0 cm long is uniformly charged and has a total charge of -20.0 µc. determine the magnitude and direction of the electri

c field along the axis of the rod at a point 32.0 cm from its center.
Physics
2 answers:
sineoko [7]2 years ago
7 0
<span>Q is the charge and it is Q = -20.0 µC.</span>
Let D denotes the the distance between the center of the rod and the point.Then,
D=0.32 - 0.12 = 0.2 m
<span>L = 0.12 m - the length of the rod
</span><span>The magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center can be obtained with the formula: 
</span><span>E = K·Q/r²
</span>E = kQ/D(D+L), where k<span> is a constant with a value of 8.99 x 10</span>9<span> N m</span>2/C2.
<span>So,
E=(</span>8.99 x 109 N m2/C2.* (-20.0 µC))/(<span>0.2 m*0.32m)</span><span>

</span>
wolverine [178]2 years ago
7 0

The formula for the electric field generated by a charged rod is:

E = k Q / [D (D + L)]

Where,

E = electric field

k = Coulomb’s constant = 9 * 10^9 N m^2 / c^2

Q = total charge = -20.0 µc = -20.0 * 10^-6 c

L = length of rod = 12.0 cm = 0.12 m

D = distance from the center = 32.0 – 12.0 cm = 0.20 m

 

Substituting the known values to the formula:

E = (9 * 10^9 N m^2 / c^2) * (-20.0 * 10^-6 c) / [0.20 m (0.20 m + 0.12 m)]

E = -2,812,500 N

Since E is negative therefore E is pointing towards the center of the rod.

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The relationship between the frequency and wavelength of a wave is given by the equation:

v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency. 

If we divide the equation by f we get:

λ=v/f

From here we see that the wavelength and frequency are inversely proportional. So as the frequency increases the wavelength decreases. 

So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.  
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Technician A says that the use of some RTV sealants to seal components on an engine can damage the oxygen sensor. Technician B s
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Answer: The correct option is C(both A and B)

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7 0
2 years ago
Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. The electric field at a point P, which
krok68 [10]

Answer:

E_{max}=41666.66\ N/C

Explanation:

Given that,

The radius of sphere, r = 0.3 m

Distance from the center of the sphere to the point P, x = 0.5 m

Electric field at point P, E_P=15000\ N/C (radially outward)

The maximum electric field is at the surface of the sphere. We know that the electric field is inversely proportional to the distance. So,

\dfrac{E_{max}}{E_p}=\dfrac{0.5^2}{0.3^2}

\dfrac{E_{max}}{15000}=\dfrac{0.5^2}{0.3^2}

{E_{max}}=\dfrac{0.5^2}{0.3^2}\times 15000

E_{max}=41666.66\ N/C

So, the magnitude of the electric field due to this sphere is 41666.66 N/C. Hence, this is the required solution.

6 0
2 years ago
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
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Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

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⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
2 years ago
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