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densk [106]
1 year ago
6

M A 40.0- cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux is found.

The flux in this position is measured to be 5.20 × 10⁵N . m²/C. What is the magnitude of the electric field?
Physics
1 answer:
Oksi-84 [34.3K]1 year ago
3 0

By electric flux, the electric field is 41.36 x 10⁵ N/C

We need to know about electric flux to solve this problem. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It can be determined as

Φ = E . S = E . Scosθ

where Φ is electric flux, E is electric field, θ is angle of surface and S is surface area.

From the question above, we know that

Φ = 5.20 × 10⁵ N.m²/C

d = 0.4 m²

r = 0.2 m²

because of maximum flux, hence (cosθ = 1)

By substituting the following parameter, we get

Φ = E . Scosθ

5.20 × 10⁵ =E  . π.r²

5.20 × 10⁵ =E . π.0.2²

E = 41.36 x 10⁵ N/C

Hence, the electric field is 41.36 x 10⁵ N/C

For more on electric flux at: brainly.com/question/26289097

#SPJ4

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A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface
Fiesta28 [93]

Answer:

X_2=25.27m

Explanation:

Here we will call:

1. E_1: The energy when the first spring is compress

2. E_2: The energy after the mass is liberated by the spring

3. E_3: The energy before the second string catch the mass

4. E_4: The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. E_1 = E_2

2. E_2 -E_3= W_f

3.E_3 = E_4

where W_f is the work of the friction.

1. equation 1 is equal to:

\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2

where K is the constant of the spring, x is the distance compressed, M is the mass and V_2 the velocity, so:

\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2

Solving for velocity, we get:

V_2 = 65.319 m/s

2. Now, equation 2 is equal to:

\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd

where M is the mass, V_2 the velocity in the situation 2, V_3 is the velocity in the situation 3, U_k is the coefficient of the friction, N the normal force and d the distance, so:

\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)

Volving for V_3, we get:

V_3 = 65.27 m/s

3. Finally, equation 3 is equal to:

\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2

where K_2 is the constant of the second spring and X_2 is the compress of the second spring, so:

\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2

solving for X_2, we get:

X_2=25.27m

3 0
3 years ago
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