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3 years ago

Why does carpet produce static electricity more than hardwood floors?

Physics

2 answers:

NeX [460]3 years ago

7 0

Answer:

The carpet has a buildup of electrons in it that get released onto your body when you rub across the floor while the hardwood doesn't have electrons built up inside of it.

Explanation:

I study that

DiKsa [7]3 years ago

4 0

Carpet produce more static electricity because the rubbing on the carpet release negative charge, called electrons, which can build up on one object to produce a static charge. For example, when you shuffle your feet across a carpet, electrons can transfer onto you, building up a static charge on your skin.

You might be interested in

Which of the following statements are true with respect to the law of conservation of mass?

Mazyrski [523]

Answer:

Only (I) is true

i) Mass is conserved during a chemical reaction.

8 0

2 years ago

A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l

hram777 [196]

Answer:

$t = 37.6 s$

Explanation:

As we know that train is initially moving with the speed

$v_i = 75.1 km/h$

now we know that

$v_i = 20.86 m/s$

now the final speed of the train when it crossed the crossing

$v_f = 15 km/h$

$v_f = 4.17 m/s$

now we can use kinematics here

$v_f^2 - v_i^2 = 2 a d$

$4.17^2 - 20.86^2 = 2 a(471)$

$a = -0.44 m/s^2$

Now the time to cross that junction is given as

$v_f - v_i = at$

$4.17 - 20.86 = a(-0.44)$

$t = 37.6 s$

4 0

2 years ago

At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t

STatiana [176]

I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!

4 0

3 years ago

Read 2 more answers

A loop of wire is placed in a uniform magnetic field. For what orientation of the loop is the magnetic flux a maximum? For what

andreyandreev [35.5K]

Answer:

The flux is calculated as φ=BAcosθ. The flux is thereforemaximum when the magnetic field vector is perpendicular to theplane of the loop. We may also deduce that the flux is zero whenthere is no component of the magnetic field that is perpendicularto the loop.

when angle is zero then flux is maximium because when angle zerocos is maximium

8 0

3 years ago

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

2 years ago