An axe used to split wood is driven into a piece of wood for an input distance of 0.05m. If the mechanical

A certain resistance thermometer read 14.5 ohms in pure melting ice and 18.5 ohms in steam at standard atmospheric pressure what

Vadim26 [7]

The resistance of the thermometer at room temperature is 15.04 ohms.

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<h3>What is a resistance thermometer?</h3>

A resistance thermometer is a type of thermometer that measures temperature through a change in resistance.

To calculate the resistance of the thermometer at room temperature, we use the formula below.

Formula:

100/27 = 2/(x-14.5)..............Eqquation 1

Where:

x = Resistance of the thermometer at room temperature

Make x the subject of the equation

x = [(27×2)/100]+14.5
x = (54/100)+14.5
x = 0.54+14.5
x = 15.04 ohms.

Hence, The resistance of the thermometer at room temperature is 15.04 ohms.

Learn more about thermometers here: brainly.com/question/1531442

3 0

2 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

So

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

So

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

3 years ago

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2

Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × $10^{5}$ molecules $cm^{-3}$ , the result will be that the reaction with OH will be 535 + (100 to about 4 × $10^{5}$ molecules $cm^{-3}$ ) times faster than the reaction with Cl

Explanation:

4 0

3 years ago

Does anyone know this one? Thanks

Inessa [10]

Answer:

$3.844\,*\,10^5$

So a=3.844 and b=5

Explanation:

Scientific notation requests to write a number using powers of ten as a factor accompanying a real number (a) between 1 and smaller than 10 that contains the digits to exactly represent the original number. So in this case, the number 384,400 can be written as:

$384,400=3.844 \,*\,100,000= 3.844 \,*\,10^5$

with a=3.844, and "5" as the exponent of ten (so b=5)

6 0

4 years ago

An Na ion has a charge of +1 and an O ion has a charge of –2. How many Na ions does one O ion need to balance charges

Aleonysh [2.5K]

Because Na⁺ ion lacks an electron and O²⁻ has two extra electrons extra, to balance the charge, we need 2 Na⁺ ions.

All ions, atoms and molecules want to get to the minimum energy state, and that state is when the ion, atom, or molecule is neutral, that's why all of them want to balance their charges.

7 0

3 years ago

Read 2 more answers