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earnstyle [38]
3 years ago
8

The rectangular region of the xy plane shown has nonuniform surface charge density σ = σ0 (1+ y b), where σ0 is a constant. Find

the net charge.
Physics
1 answer:
sergiy2304 [10]3 years ago
8 0

Answer:

This is net charge on the surface  is  Q = σ₀ x (y + 2by²)

Explanation:

The surface charge density is defined as the amount of charge Q per unit area A

       σ = dq / dA

       dq = σ dA

Since the surface is a rectangular region we use an xy coordinate system so the area difference  

      dA = dxdy

      dq = σ dx dy

 We replace, evaluate the integral

        ∫ dq = ∫ σ₀ (1 + yb) dxdy

realizamos laintegral de dx

        Q -0 =σ₀ ∫ (1 + yb) (x-0)   dy

Where we evaluate We must recognize that the charge Q must be zero by the time X = 0 and Y = 0. At the starting point Q = 0 for x = 0

 

We perform the other integral (dy)

        Q = σ₀ x (y + 2y² b)

Evaluated between Y = 0 and Y = y

      Q = σ₀ x (y + 2by²)

This is net charge on the surface

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frutty [35]

Newton's subsequent law expresses that power is corresponding to what exactly is needed for an object of consistent mass to change its speed. This is equivalent to that item's mass increased by its speed increase.

We use Newtons, kilograms, and meters each second squared as our default units, albeit any proper units for mass (grams, ounces, and so forth) or speed (miles each hour out of every second, millimeters per second², and so on) could unquestionably be utilized also - the estimation is the equivalent notwithstanding.

Hence, the appropriate answer will be 399,532.

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7 0
3 years ago
A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
Veronika [31]

Answer:

14.98\ \text{kg m/s}

45.26^{\circ}

Explanation:

P_1 = Initial momentum of the pin = 13 kg m/s

P_i = Initial momentum of the ball = 18 kg m/s

P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

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34kurt
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