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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.

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Consider a rifle, which has a mass of 2.44 kg and a bullet which has a mass of 150 grams and is loaded in the firing chamber.Whe

svetlana [45]

Answer:

a. 0 kgm/s

b. 0 kgm/s

c. 66 kgm/s

d. -66 kgm/s

e. 0 kgm/s

f. -27.05 m/s

g. 173.68 N

h. 12.58 m/s

i. 0.772 m

j. 14487 J

Explanation:

150 g = 0.15 kg

a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.

b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.

c. After the bullet is fired, the momentum is:

0.15*440 = 66 kgm/s

d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s

e. 0 according to law of momentum conservation.

f. Velocity of the rifle is its momentum divided by mass

v = -66 / 2.44 = -27.05 m/s

g. The average force would be the momentum divided by the time

f = -66 / 0.38 = 173.68 N

h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is

66 / 5.25 = 12.58 m/s

i. The normal force and also friction force due to sliding is

$F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N$

According to law of energy conservation, the initial kinetic energy will soon be transformed to work done by this friction force along a distance d:

$W = K_e$

$dF_f = 0.5Mv_0^2$

$d = \frac{Mv_0^2}{2F_f} = \frac{5.25*12.58^2}{2*42.75} = 0.772 m$

j.Kinetic energy of the bullet before the impact:

$K_b = 0.5*m_bv_b^2 = 0.5*0.15*440^2 = 14520 J$

Kinetic energy of the block-bullet system after the impact:

$K_e = 0.5Mv_0^2 = 0.5 * 5.25*12.58^2 = 33 J$

So 14520 - 33 = 14487 J was lost during the lodging process.

3 0

3 years ago

Motion to the right is given a _________ sign, and motion to the left is given a negative __________.

siniylev [52]

Answer: Motion to the right is given a negative_________ sign, and motion to the left is given a negative velocity function __________.

Explanation:

5 0

3 years ago

A point charge that is exactly q =3 mu or micro CC is at the origin. In this problem, assume that the Coulomb constant k = 8.99

Simora [160]

Answer:

(a) V1 = 8990.00 V

V2 = 8960.13 V

Explanation:

Parameters given:

q =3 mC

k = 8.99 * 10⁹ Nm²/C²

x1 = 3 m

x2 = 3.01 m

Electric potential is given as:

V = kq/r

Where

k = Coulombs constant

q = charge

r = distance

Potential at x1 is:

V1 = (8.99 * 10⁹ * 0.000003)/(3)

V1 = 8990.00V

Potential at x2 is:

V2 = (8.99 * 10⁹ * 0.003)/(3.01)

V2 = 8960.13 V

7 0

3 years ago

Read 2 more answers

Two protons in a molecule are 4.40 10-10 m apart. Find the electric force exerted by one proton on the other.

Akimi4 [234]

Answer:

F = 1.19 x 10⁹ N

Explanation:

The electrostatic force between the charged particles is given by Colomb's Law as follows:

$F = \frac{kq_{1}q_{2}}{r^{2}}$

where,

F = Electrostatic force = ?

k = Colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = q₂ = charge on proton = 1.6 x 10⁻¹⁹ C

r = distance between protons = 4.4 x 10⁻¹⁰ m

Therefore,

$F = \frac{(9\ x\ 10^{9}\ Nm^{2}/C^{2})(1.6\ x\ 10^{-19}\ C)(1.6\ x\ 10^{-19}\ C)}{(4.4\ x\ 10^{-10})^{2}} \\\\$

8 0

3 years ago

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0

3 years ago