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2 years ago

Where are you most likely to build up enough static charge to receive a

2 answers:

6 0

You're most likely to build up enough static charge to receive a shock by walking around in a carpeted restaurant in the desert. (A)

Walking on carpet is the fastest way to accumulate charge, and the dry desert air prevents the charge from dribbling off of you and away.

When I walked on stones in the Sinai Desert, the dry wind with a little bit of sand or dust in it built up enough static charge on me that I got a shock every time I stood less than a foot away from my partner.

I had the same experience a few years later near Ouarzazate in the interior of Morocco.

When you hear people say "the desert is dry", they mean it's <em>DRY ! </em>

gulaghasi [49]2 years ago

4 0

Answer:

A. In a carpeted resteraunt in the desert

Explanation:

i like anime

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The primary service used by stations to exchange mac frames when the frame must traverse the ds to get from a station in one bss

Alenkinab [10]

<span>The primary service used by stations to exchange mac frames when the frame must traverse the ds to get from a station in one bss to a station in another bss is Distribution
In this type of service, the bss is required in order to build the basic building block for the wireless LAN</span>

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3 years ago

Explain about ohm's law.

belka [17]

Answer:

Statement:

The electric current passing through a conductor is directly proportional to the potential difference across its ends provided temperature and other physical conditions remain constant.

Explanation:

Current is directly proportional to voltage loss through a resistor. That is, if the current doubles, then so does the voltage. To make a current flow through a resistance there must be a voltage across that resistance. Ohm's Law shows the relationship between the voltage (V), current (I) and resistance (R).

V∝I or I∝V⇒V=IR.

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3 years ago

Read 2 more answers

A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin return

tensa zangetsu [6.8K]

Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2(0.84)

T = 1.68 s.

T = 1.68 s

7 0

3 years ago

How would you describe the appearance of the substance after the phase change?

Leona [35]

Answer:

Substances can change phase—often because of a temperature change. At low temperatures, most substances are solid; as the temperature increases, they become liquid; at higher temperatures still, they become gaseous. The process of a solid becoming a liquid is called melting.

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3 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

7 0

3 years ago