If a charge is located at the center of a spherical volume and the electric flux through the surface of the sphere is φ o, what is the flux through the surface if the radius of the sphere doubles?

Answers

Answer 1

Answer:

The electric flux remains unchanged

Explanation:

From Gauss law the Electric flux is directly proportional to the number of electric field lines passing through a surface. The number of field lines passing through a surface become if the radius is doubled becomes 1/4th that is when radius of the Gaussian surface is doubled, but at the same time, the surface area has increased 4 times , so the electric flux remains unchanged

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a force of 35N is exerted over a cylinder with an area of 5m^2. What pressure,in pascals, will be transmitted in the hydraulic system?

Answers

Answer:

The answer is 7 Pa

Explanation:

The pressure transmitted in the hydraulic system can be found by using the formula

$p = (f)/(a) \n$

f is the force

a is the area

From the question we have

$p = (35)/(5) \n$

We have the final answer as

7 Pa

Hope this helps you

The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally from west to east. the vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 3.00 10-5 t. (a) in what direction are the electrons deflected by this field component? due north due south due east due west (b) what is the magnitude of the acceleration of an electron in part (a)? m/s2

Answers

(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
$K= (1)/(2)mv^2$
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
$v= \sqrt{ (2K)/(m) }= \sqrt{ (2 \cdot 2.20 \cdot 10^(-15) J)/(9.1 \cdot 10^(-31) kg) }=6.95 \cdot 10^7 m/s$

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
$qvB = m (v^2)/(r)$
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
$r= (mv)/(qB)= ((9.1 \cdot 10^(-31) kg)(6.95 \cdot 10^7 m/s))/((1.6 \cdot 10^(-19) C)(3.00 \cdot 10^(-5) T))=13.18 m$

And finally we can calculate the centripetal acceleration, given by:
$a_c = (v^2)/(r)= ((6.95 \cdot 10^7 m/s)^2)/(13.18 m)=3.66 \cdot 10^(14) m/s^2$

A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pull the bucket of water upwards. the total mass of the bucket of water is f2= 3.9kg. -Calculate how much work Wg in J gravity does on the bucket filled with water as the farmer lifts it up the well.
-Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.

Answers

Final answer:

To calculate the work done by gravity on the bucket of water as it is lifted up the well, multiply the weight of the bucket by the lifting distance. The net work done on the bucket by the force applied by the farmer and gravity is the sum of the work done by both forces. The net work is represented by the equation Wnet = W1 + Wg.

Explanation:

To calculate how much work gravity does on the bucket filled with water as the farmer lifts it up the well, we need to multiply the force of gravity (weight) by the vertical distance the bucket is lifted. The equation for work is W = Fd, where W is the work done, F is the force, and d is the distance. In this case, the force of gravity is the weight of the bucket, which can be calculated by multiplying the mass by the acceleration due to gravity (9.8 m/s^2).

So, the work done by gravity (Wg) on the bucket is Wg = Fg * d = (m * g) * d = (3.9 kg * 9.8 m/s^2) * d = 38.22 d Joules.

To calculate the net work done on the bucket by the two forces, we can use the equation Wnet = W1 + Wg, where W1 is the work done by force F1 and Wg is the work done by gravity. Since force F1 and the displacement (lifting distance) are both vertical, the work done by F1 is given by W1 = F1 * d.

Therefore, the net work done on the bucket by forces F1 and gravity is Wnet = F1 * d + Fg * d = (57.5 N) * d + (3.9 kg * 9.8 m/s^2) * d = (57.5 N + 38.22 d) Joules.

Learn more about Work done by forces here:

brainly.com/question/31428590

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Swinging a tennis racket against a ball is an example of a third class lever.
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Please select the best answer from the choices provided.
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Answers

Final answer:

Swinging a tennis racket against a ball as a third class lever in physics.

Explanation:

Swinging a Tennis Racket as a Third Class Lever

A tennis racket swinging against a ball is indeed an example of a third class lever in physics. In a third class lever, the effort is situated between the fulcrum and the load. In this case, the effort is provided by the player's hand gripping the racket handle, the fulcrum is the wrist joint, and the load is the ball being struck by the racket.

When a player swings the racket, the force applied by the player's hand exerts an effort on the handle of the racket. This causes the racket to rotate about the wrist joint acting as the fulcrum. The ball serves as the load, receiving the force and accelerating in the opposite direction.

Learn more about Third class lever here:

brainly.com/question/4532561

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Answers

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=(1)/(2)kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_(i)=E_(f)$

$U_(i)+U'_(i)=U_(f)+U'_(f)$

$(1)/(2)kx^2+0=0+mgh$

$h=(kx^2)/(2mg)$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=(5365*(0.097)^2)/(2*0.221*9.8)$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

A 500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. If the total energy of the system is 4J, find the force constant of the spring?