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Sunny_sXe [5.5K]
1 year ago
6

Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr

ic field at a point between the two surfaces?
Physics
1 answer:
svlad2 [7]1 year ago
6 0

Answer:

Electric field, E = 45.19 N/C

Explanation:

It is given that,

Surface charge density of first surface, \sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2

Surface charge density of second surface, \sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2

The electric field at a point between the two surfaces is given by :

E=\dfrac{\sigma}{2\epsilon_o}

E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}

E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}

E = 45.19 N/C

So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.

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A

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You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill
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Answer:

4.325\times10^6J

Explanation:

Mass of the cable car, m = 5800 kg

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Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = \frac{1}{2} mv^2). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = 5800\times9.81\times76.0166J=4.325\times10^6J

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The seeds were sown (change the voice)​
MaRussiya [10]

Answer:

He sowed the seeds

Explanation:

While the question should have given the person who did the sowing for example the seeds were sown by him/ her/ the farmer/ or any name. Therefore, voice is given in passive and to change passive voice to active voice then the sentence will read as follows assuming that the seeds were sown by him

He sowed the seeds

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Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass
zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value into the formula

8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v

v=\dfrac{-19.375}{14.25}

v=-1.35\ m/s

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2

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We need to calculate the final kinetic energy

Using formula of kinetic energy

K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2

Put the value into the formula

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We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

\Delta K.E=K.E_{f}-K.E_{i}

Put the value into the formula

\Delta K.E=12.98-239.96

\Delta K.E=-226.98\ J

Negative sign shows the loss of mechanical energy

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(b). The mechanical energy dissipates during this play is 226.98 J.

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