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2 years ago

What type of charge is acquired when a rubber is rubbed with a fur

Physics

1 answer:

Rus_ich [418]2 years ago

4 0

Answer:

The triboelectic charging process (a.k.a., charging by friction) results in a transfer of electrons between the two objects that are rubbed together. Rubber has a much greater attraction for electrons than animal fur.

Explanation:

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A force of 30 N is applied tangentially to the rim of a solid disk of radius 0.14 m. The disk rotates about an axis through its

jenyasd209 [6]

Answer:

3.8 kg

Explanation:

The torque, τ, due to the force, F, is given by

$\tau = F\times r$

where r is the radius.

This torque is also given by

$\tau = I\times\alpha$

where I and α are respectively the moment of inertia and the angular acceleration.

For a solid disk, its moment of inertia for an axis though its centre and perpendicular to its face is given by

$I = \frac{1}{2}mr^2$

where m is its mass and r is its radius.

Hence, we have

$\tau = F\times r = I\times\alpha=\frac{1}{2}mr^2\times\alpha$

$m = \dfrac{2F}{r\alpha} = \dfrac{2(30\ \text{N})}{(0.14\ \text{m})(115\ \text{rad/s}^2)} = 3.8\ \text{kg}$

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3 years ago

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||||| A ball on a porch rolls 60 cm to the porch’s edge, drops 40 cm, continues rolling on the grass, and eventually stops 80 cm

3241004551 [841]

Answer:

$d = 145.6 cm$

Explanation:

first ball rolls on the porch by total distance

$x_1 = 60 cm$

Then again it will move on horizontal floor

$x_2 = 80 cm$

also in vertical direction it will drop down

$y = 40 cm$

so we have

$x = 60 + 80$

$x = 140 cm$

$y = 40 cm$

so magnitude of net displacement of the ball is given as

$d = \sqrt{140^2 + 40^2}$

$d = 145.6 cm$

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3 years ago

True or False? Sometimes, substances can be added to a reaction to cause the reaction to start or speed up. These are not consid

Sphinxa [80]

The correct anwser is A true

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4 years ago

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A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

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3 years ago

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