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Umnica [9.8K]
3 years ago
5

A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field \vec{E} = (100 N/C) \ha

t{i}E → = ( 100 N / C ) i ^. Calculate the flux of the net electric field through a Gaussian sphere of radius R = 10.0 \text{ cm}R = 10.0 cm centered at the origin. (ε0 = 8.85 × 10-12 C2/N ∙ m2)
Physics
1 answer:
tester [92]3 years ago
8 0

Answer:

Φ = 2.26 10⁶ N m² / C

Explanation:

The electric flow is

           Ф = E .ds = q_{int} / ε₀

Since we have two components for this flow. The uniform electric field and the flux created by the point charge.

The flux of the uniform electric field is zero since the flux entering is equal to the flux leaving the Gaussian surface

 

The flow created by the point load at the origin is

              Φ = q_{int} /ε₀

Let's calculate

              Φ = 2.00 10⁻⁶ /8.85 10¹²

              Φ = 2.26 10⁶ N m² / C

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2 years ago
A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
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Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

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Voltage needed to raise current to 3.75a using 20,20,200 resistor set
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<u>Answer:</u> The voltage needed is 35.7 V

<u>Explanation:</u>

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

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We are given:

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Using above equation, we get:

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where,

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Putting values in equation 1, we get:

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