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Effectus [21]
3 years ago
5

Find the electric flux through a cylindrical Guassian surface produced by a point charge at base.

Physics
1 answer:
Simora [160]3 years ago
7 0

Answer:

Electric flux will be \Phi =\frac{q}{\epsilon _0}

Explanation:

Let the point charge enclosed by Gaussian surface is q

We have to find the electric flux through this Gaussian surface

According to Gauss law electric flux through any closed surface is equal to \frac{1}{\epsilon _0} times of charged enclosed by the surface.

So electric flux will be \Phi =\frac{q}{\epsilon _0}

Gauss law is only valid for closed surface it is not applied for open surface

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Which of the following material offers the lowest resistivity?
Yuki888 [10]
Lower resistivity means higher conductivity: among these materials, the best conductor is silver, therefore it must be the material with lowest resistivity among those. We can also have a look at the value of resistivity of the different materials listed in the problem, to check our answer. The resistivities of the materials are the following:
Wood: 1.00\cdot 10^{3} \Omega \cdot m
Salt water: 2.00\cdot 10^{-1} \Omega \cdot m
Silver: 1.59\cdot 10^{-8} \Omega \cdot m
Lead: 2.20\cdot 10^{-7} \Omega \cdot m
<span>We see that the material with lowest resistivity among those is silver, therefore the correct answer is silver.</span>
7 0
3 years ago
Read 2 more answers
Match each label to the boundary it describes. convergent boundary new crust forms transform boundary crust submerges into the m
Paraphin [41]

The answers would be:

CONVERGENT boundary - Crust submerges into the mantle

TRANSFORM boundary - neither forms nor submerges

DIVERGENT boundary - new crust forms

If you'd like to know more about the different boundaries, read on:

Convergent boundaries occur when two plates move TOWARDS each other. The event where crust submerges into the mantle is called <em><u>subduction</u></em> and this occurs when an oceanic plate and a continental plate collide. The oceanic plate is more dense and thinner than the continental plate, so it slides under it.

Transform boundaries occur when two plates slide against each other. They move slide side by side, so nothing is formed nor do they go under each other. Although, this type of boundaries create strong earthquakes.

Lastly, divergent boundaries occur when two plates move apart. The separation creates a way for magma to come up. New crust is formed when the magma that seeps out is cooled by its cooler surroundings. This is observed in the mid oceanic ridge.

7 0
3 years ago
Read 2 more answers
Projectiles Launched Horizontally Quiz
NikAS [45]

Answer:

3.0 seconds

Explanation:

The time of flight of a projectile (the time it takes to reach the ground) does not depend on the horizontal motion, but only on its vertical motion.

In fact, the time of flight is determined by the suvat equation:

s=ut+\frac{1}{2}gt^2

where

s is the vertical displacement

u is the initial vertical velocity (0, in case of these two projectiles)

g = 9.8 m/s^2 is the acceleration of gravity (assuming downward as positive direction)

t is the time of flight

Re-arranging the equation, we get

t=\sqrt{\frac{2s}{g}}

We see that this time depends only on s (the heigth of the cliff) and g: therefore, since the two projectiles are launched from the same height, they take the same time to reach the ground, 3.0 seconds.

7 0
3 years ago
Two boys with masses of 40 kg and 60 kg stand on a horizontal frictionless surface holding the ends of a light 10-m long rod. Th
kari74 [83]

When they meet the 40-kg boy will have moved a distance of 6 m.

Displacement of the 40 kg boy

The displacement of the 40 kg boy is calculated from the principle of center mass.

X(40 kg) = (60 x 10 m    +  40 x 0)/(40 kg + 60 kg)

X(40 kg) = (600)/(100) = 6 m

X(60 kg) = (60 x 0    +  40 x 10 m)/(40 kg + 60 kg)

X(60 kg) = (400)/(100) = 4 m

Thus, when they meet the 40-kg boy will have moved a distance of 6 m.

Learn more about center mass here: brainly.com/question/13499822

#SPJ1

7 0
2 years ago
A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
3 years ago
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