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Aleonysh [2.5K]
1 year 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]1 year 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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A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart
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When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:

F = mg

Where,

m = mass

g = Gravitational acceleration

F = 5*9.8

F = 49N

Therefore the correct answer is E.

5 0
3 years ago
In your own words explain coefficient of friction?​
fomenos

Answer:

A coefficient friction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.

Explanation:

3 0
3 years ago
a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^
posledela

Answer:

Approximately 0.8\; \rm s (assuming that air resistance is negligible.)

Explanation:

Let v_0 denote the initial velocity of this ball. Let \theta denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

  • Initial vertical velocity: v_0(\text{vertical}) = v_0 \cdot \sin(\theta).
  • Initial horizontal velocity: v_0(\text{vertical}) = v_0 \cdot \cos(\theta).

If air resistance on this ball is negligible, v_0(\text{vertical}) alone would be sufficient for finding the time of flight of this ball.

Calculate v_0(\text{vertical}) given that v_0 = 8 \; \rm m \cdot s^{-1} and \theta = 30^\circ:

\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}.

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}, the vertical velocity of this ball right before landing would be v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}.

Calculate the change to the vertical velocity of this ball:

\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}.

In other words, the vertical velocity of this ball should have change by 8\; \rm m \cdot s^{-1} during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is g = 10\; \rm m \cdot s^{-2}. In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by 10\; \rm m\cdot s^{-1} each second. What fraction of a second would it take to change the vertical velocity of this ball by 8\; \rm m \cdot s^{-1}?

\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}.

In other words, it would take 0.8\; \rm s to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be 0.8\; \rm s\!.

5 0
3 years ago
If the mass of a material is 49 grams and the volume of the material is 10 cm^3, what would the density of the material be?
Irina-Kira [14]
Answer: 4.9 g/cm^3

Explanation.

!) Data

mass = 49 g
Volume = 10 cm^3

2) Formula

Density = mass / volume

Density = 49 g / 10cm^3 = 4.9 g/cm^3

This is a problem where you only need to know the formula and subtitute the numbers to calculate.

The formula of density is the very same definition: mass per unit of volume.

So, as you had the mass and the volume you just need to plug the values in the formula density = mass / volume.
8 0
3 years ago
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