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Aleonysh [2.5K]
2 years ago
15

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the tot

al flux leaving the cylinder, what portion of the flux leaves the curved surface of the cylinder
Physics
1 answer:
zheka24 [161]2 years ago
5 0

The portion of the flux leaves the curved surface of the cylinder is 60%.

<h3 /><h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.

If 20% of the flux leave from one end, then another 20% will leave from another end.

So, the net flux through curved surface is

100 -20 -20 = 60%

Thus, the total flux  leaves the curved surface of the cylinder is 60%

Learn more about electrons.

brainly.com/question/1255220

#SPJ1

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Answer: The galaxy's bulge

Explanation:

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The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by
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Here we can use momentum conservation as in this type of collision there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}

now here we can say

m_1 = 10 g

v_{1i} = 0

m_2 = 550 g

v_{2i} = 3.5 m/s

now here we can say

10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}

192.5 = v_{1f} + 55 v_{2f}

now by coefficient of restitution

for elastic collision we know that e = 1

v_{2f} - v_{1f} = e(v_{1i} - v_{2i})

v_{2f} - v_{1f} = 0 - 3.5

now by solving the two equation

56v_{2f} = 189

v_{2f} = 3.375 m/s

also we know that

v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s

so final speed of the nail is 6.875 m/s


6 0
3 years ago
Read 2 more answers
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

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3 years ago
What is the primary evidence used to determine how the Moon formed? O A. Moon craters and Earth craters were caused by the same
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A farsighted boy has a near point at 2.3 m and requires eyeglasses to correct his vision. Corrective lenses are available in inc
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Answer:

P = 3.5 D

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As we know that convex lens is to be used to make the near point of eye to be correct

So we will have

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here we have

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now plug in all values into the formula

-\frac{1}{230} + \frac{1}{25} = \frac{1}{f}

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now for power of lens

P = \frac{1}{f}

P = \frac{1}{0.28} = 3.5 D

so the power in dioptre is

P = 3.5 D

5 0
3 years ago
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