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2 years ago

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A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax

es of a rectangular coordinate system and the corner formed by these two sides is at the origin. An electric field of magnitude E=bx2 directed along the +x axis exists in this region. How much charge is contained in the volume of the shell?

1 answer:

Bingel [31]2 years ago

5 0

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

Φ = EA

E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

so that

Q = ba⁴ * ε₀

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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

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2 years ago

An astronaut holds a rock 100m above the surface of Planet X . The rock is then thrown upward with a speed of 15m/s , as shown i

Butoxors [25]

Answer: $5 m/s^{2}$

Explanation:

The described situation is is related to vertical motion (and free fall). So, we can use the following equation that models what happens with this rock:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Then, isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

Finally:

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

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1 year ago

In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li

Alenkinab [10]

Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

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So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

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We know that energy $E=power\times time=6.8\times 3=20.4KWH$

Now power of each CFL bulb = 25 W

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vlabodo [156]

Answer:

24.3 degrees

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Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

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$tan\alpha = 0.451$

$\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0$

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2 years ago