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never [62]
4 years ago
10

Hi, I'm having a lot of trouble with this one. The answer I got to was 407.91 but I'm not confident in it.

Physics
1 answer:
Alecsey [184]4 years ago
3 0

Answer:

6.20×10⁴ V/m

Explanation:

The magnitude of electric field is:

E = √(Eₓ² + Eᵧ²)

where Eₓ = ∂φ/∂x and Eᵧ = ∂φ/∂y.

φ = 1.11 (x² + y²)^-½ − 429x

Eₓ = -0.555 (x² + y²)^-(³/₂) (2x) − 429

Eᵧ = -0.555 (x² + y²)^-(³/₂) (2y)

Evaluating at (0.003, 0.003):

Eₓ = -44034 V/m

Eᵧ = -43605 V/m

The magnitude is:

E = 61971 V/m

Rounded to three significant figures, the strength of the electric field is 6.20×10⁴ V/m.

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Match each scenario to the form of energy it represents.
Pachacha [2.7K]
<span>Correct pairs:

a man jogging in the park --> motion energy (the energy is the kinetic energy of the man, moving with speed v)

a fully charged camera battery --> electric potential energy (the battery is fully charged, so it can deliver electrical energy when the camera is turned on)

a stove burner that’s turned on --> radiant energy (the stove burner emits energy by radiation)

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5 0
3 years ago
will the value of g be affected by the radius of the earth? consider the real shape of the earth. compare the acceleration due t
sp2606 [1]

Answer:

Yes, the value of g affected by the radius.

Explanation:

The formula for the force of gravity of 2 objects is

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4 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

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Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

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C_d=0.45

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Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

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We need calculate now the energy lost through a time T, then,

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v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

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Alja [10]
Hope this helps you!

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