Which statement is true regarding DC current?

Given that the atomic weight of hydrogen is approximately 1 gram per mole, use the method above to estimate the average molecula

denpristay [2]

Answer:

2

Explanation:

6 0

3 years ago

How do you think overpumping groudwater is related to the formation of sinkholes?

vladimir2022 [97]

Ground water keept the ground at a stable level when it is gone the cavern it was in has no support and is at risk of callaps

3 0

4 years ago

If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock com

Lelu [443]

Answer:

<em> B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.</em>

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = $100*10^{-6} R$

for dry skin, voltage will be

V = IR = $100*10^{-6}*200R = 0.02R$

Comparing both voltages

$0.02R$ ÷ $100*10^{-6} R$ = 200

<em>this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.</em>

4 0

4 years ago

What method has detected the most extrasolar planets so far?

Sedaia [141]

<h2>Answer: The Transit method</h2>

Detecting extrasolar planets by direct observation (with a telescope) is a complicated task. This is because any planet constitutes an extremely dim light source compared to the star around which it orbits.

So, to detect this extremely dim source is quite difficult due to the glare of the star's light that dulls it.

In this sense, scientists and astronomers have made several methods to find these extrasolar planets, among which the most successful has been the transit method.

This method is based on <u>astronomical transit</u>, a phenomenon in which a body (a planet in this case) passes in front of a larger one (the star), blocking (eclipsing) its vision to some extent.

It should be noted that this is the method currently used in the search for extrasolar planets. Space agencies such as ESA (Europe) and NASA (USA) have put into orbit satellites with extremely sensitive photometric sensors to observe even the smallest variations of intensity of a star due to the passage of a planet.

5 0

3 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

5 0

3 years ago