A Н20= C=1 h=4 o=4 2. H3PO4 + КОН KзРО І

Nerve signals in the body occur when a small voltage, called an action potential, is applied across the membrane of a cell. When

nasty-shy [4]

Answer: 23.56 nV

Explanation:

Assuming that we apply Ohm's Law to this situation, we know that under this condition, the current resultant, is proportional to the voltage applied, and that the proportionality constant, is called the resistance.

Now, we define electric current, as the passage of a number of charges over time.

In this case, we can say that each singly-ionized potassium ion carries the charge equivalent to the one electron, which is q = 1.6. 10⁻¹⁹ coulombs.

So, as we have 90,000 ions, we will have a total charge as follows:

Q =90.10³. 1.6.10⁻¹⁹ coulombs

Also, we are told that this charge will be moving for 1.1 msec, so we can find the current I as follows:

I = Q/t ⇒ I = 13.09. 10⁻¹² A.

If we know that R= 1.8.10⁹ Ω, we can determine V, applying Ohm's Law, as follows:

V = 13.09. 10⁻¹² A . 1.8.10⁹ Ω = 23.56 nV

3 0

3 years ago

Read 2 more answers

Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s

tatyana61 [14]

Answer:

$Q_T=63313.5\ J$

Explanation:

Given:

temperature of skin, $T_s=34^{\circ}C$

initial temperature of steam vapour, $T_v=100^{\circ}C$

latent heat of steam, $L=2256\ J.g^{-1}$

mass of steam, $m=25\ g$

specific heat of water, $c=4190\ J.kg^{-1}.K^{-1}=4.19\ J.g^{-1}.K^{-1}$

final temperature, $T_f=34^{\circ}C$

Assuming that no heat is lost in the surrounding.

We know:

$Q=m.c.\Delta T$

Now the total heat given by the steam to form water at the given conditions:

$Q_T=Q_{Lv}+Q_w$ ..............................(1)

where:

$Q_{Lv}=$ latent heat given out by vapour to form water of 100°C

$Q_w=$ heat given by water of 100°C to come at 34°C.

putting respective values in eq. (1)

$Q_T=m(L+c.\Delta T)$

$Q_T=25(2256+4.19\times 66)$

$Q_T=63313.5\ J$

is the heat transferred to the skin.

4 0

3 years ago

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140

max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

$u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\$

Using formula:

$v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\$

$= 0.55 \ \frac{m}{sec^2}\\\\$

$F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\$

Calculating the Work by net force

$W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\$

The above work is converted into thermal energy.

Now,

$F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J$