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laila [671]
3 years ago
9

An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z

= 0 and the opposite corner at the point x=L, y=L, z=L. The cube is in a region of uniform electric field E⃗ =E1i^+E2j^, where E1 and E2 are positive constants. Calculate the electric flux through the cube face in the plane x = 0 and the cube face in the plane x=L. For each face the normal points out of the cube.

Physics
1 answer:
kodGreya [7K]3 years ago
7 0

Find the given attachment for solution

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What does the area under the velocity-time graph represent.
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Answer:

The distance traveled!

Explanation:

This is a velocity time graph of an object moving in a straight line due North.

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Sound travels through water at a speed of 1500 m/s. If the frequency of a sound is 1000 Hz, what is the wavelength?
ratelena [41]

Answer:

1.5m

Explanation:

Velocity=1500m/s

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Wavelength =velocity ➗ frequency

wavelength =1500 ➗ 1000

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On a violin, the highest note Mandy can play is an A-note, which produces a sound wave with a high frequency. The lowest note sh
lora16 [44]
I think the answer would be: The G-note's wavelength is longer

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Wavelength = Wave speed/Frequency

Which indicates that the wavelength will become larger as the frequency became smaller.

6 0
3 years ago
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At what speed does a 1,248 kg compact car have the same kinetic energy as a 18,777 kg truck going 26 km/h?
notka56 [123]

The speed of car is 100.8km/h

KE =  \frac{1}{2} m(v \: truck ) {}^{2}

= 0.5 \times 18777 \times  \frac{26}{3.6} \times  \frac{26}{3.6}

= 489708.79j

=  \frac{1}{2}  \times 1248 \times( v \: car) {}^{2}

= 624v {}^{2}

so \: v {}^{2}  =  \frac{489708.7}{624}

= 784

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=100.8km/h

Hence, the speed of the car is 100.8km/h

learn more about speed from here:

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3 0
1 year ago
Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista
soldi70 [24.7K]

Answer:

v = 7793150 m/s

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

v = \frac{kQ}{r}

                 Where     k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }

and it is satisfy the superposition principle, thus

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}

v = 222.73v

The electrical potential at 10 cm from charge 1 is:

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}

v = 395.44 v

Since the work - energy theorem, we have:

q\Delta v = \frac{mv^{2} }{2}

                     where q is the electron's charge and m is the electron's mass

Therefore:

v = \sqrt{\frac{2q\Delta v}{m} }

v = 7793150 m/s

6 0
3 years ago
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