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marysya [2.9K]
3 years ago
13

why does static electricity, the buildup of charge on a material's surface, occur in insulating material

Physics
1 answer:
likoan [24]3 years ago
7 0
When rubbing two insulators together, electrons from one insulator will be transfered to the other. so the insulator that has just received electrons now has a negative charge as it has received more electrons (making it more negative) the insulator that has just lost electrons has just become positively charged. NEVER SAY THAT PROTONS HAVE BEEN TRANSEFERD.... EVER. ONLY ELECTORNS ARE TRANSFERED.
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8 0
3 years ago
A spider spins a web with silk threads of density 1300 kg/m3 and diameter 3.0 μm . a typical tension in the radial threads of su
Tema [17]

Answer:

Explanation:

The velocity of a wave in a string is equal to:

v = √(T / (m/L))

where T is the tension and m/L is the mass per length.

To find the mass per length, we need to find the cross-sectional area of the thread.

A = πr² = π/4 d²

A = π (3.0×10⁻⁶ m)²

A = 2.83×10⁻¹¹ m²

So the mass per length is:

m/L = ρA

m/L = (1300 kg/m³) (2.83×10⁻¹¹ m²)

m/L = 3.68×10⁻⁸ kg/m

So the wave velocity is:

v = √(T / (m/L))

v = √(7.0×10⁻³ N / (3.68×10⁻⁸ kg/m))

v ≈ 440 m/s

The speed of sound in air at sea level is around 340 m/s.  So the spider will feel the vibration in the thread before it hears the sound.

5 0
3 years ago
Upon reaching a velocity of 100fps, the pilot of the airplane decides to abort the take off and applies brakes and stops the air
masha68 [24]

Answer:

5 ft/s²

Explanation:

u = Initial velocity = 100 ft/s

v = Final velocity = 0

s = Displacement = 1000 ft

a = Acceleration

From equation of motion

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-100^2}{2\times 1000}\\\Rightarrow a=-5\ ft/s^2

1\ ft/s^2=0.3048\ m/s^2\\\Rightarrow 5\ ft/s^2=5\times 0.3048\ m/s^2=1.524\ m/s^2

The airplane's deceleration is 5 ft/s² or 1.524 m/s²

8 0
3 years ago
What is the fundamental frequency of a particular medium ?
VLD [36.1K]
From my research, the question has the following choices:

a.the lowest frequency at which a standing wave is possible
b. the highest frequency at which s standing wave is possiblec. the only frequency at which a standing wave is possible
d. the only frequency at which standing wave is not possible

From there, the correct answer is A.
5 0
4 years ago
Read 2 more answers
A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is
OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

U = \frac{GMm}{r}

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}

U_2= - 5.1180*10^9J

Point B )

We now use the average radius distance from the earth.

U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}

U_2= -5.8925*10^9J

Then,

\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)

\Delta U = 774.5*10^6

By the law of conservation of energy we know that,

\Delta U = \frac{1}{2}mv^2

clearing v,

v= \sqrt{2 \Delta U/m}

v= \sqrt{2*774.5*10^6 /94.2}

v= 4055.08m/s

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0
3 years ago
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