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3 years ago

15. A locomotive moved 18.0 m [W] in a time of 6.00 s and stopped. After stopping, the

Physics

1 answer:

mariarad [96]3 years ago

3 0

Answer:

Distance = 30m

Displacement = 6m W

Explanation:

Given the following:

Movement 1 = 18m W

Movement 2 = 12m E

Diatance is a scalar quantity with only magnitude and no direction. That is, in Calculating the distance moved by the locomotive, the direction of travel or movement of the object is not considered. It only measures the total amount of movement made during the Time of motion.

Therefore, total distance traveled equals :

Movement 1 + movement 2

18m + 12m = 30m

B) Displacement also measures the movement made by an object. However, Displacement is a vector quantity and therefore, considers both magnitude and direction of travel of the object. Therefore, it measures the overall change in position of the object from its starting position.

Therefore, Displacement of the locomotive equals:

18m W - 12m E = 6m E

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How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenl

spin [16.1K]

Answer:

L=0.654 m

Explanation:

Concepts and Principles

1- The speed of sound in air is expressed as a function of the temperature of air as follows:

v=(331 m/s)√(1+T_C/273°C) (1)

where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.

Standing Wave Patterns in Pipes:

A pipe open at both ends can have standing wave patterns with resonant frequencies:

f=v/λ=nv/2L n=1,2,3.........

where v is the speed of sound in air.

Given Data

f_1 (fundamental frequency of the flute) = 262 Hz

T (temperature of the air) = 20°C

The flute is open at both ends.

Required Data

We are asked to determine the length of the tube.

Solution

The speed of sound in air at temperature T = 20°C is found from Equation (1):

v=(331 m/s)√(1+T_C/273°C)

=342.91 m/s

The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):

f=v/2L

Solve for L:

L=v/2f_1

L=0.654 m

7 0

3 years ago

Narysuj wykres zależności szybkości od czasu i drogi od czasu jeśli ciało porusza się ruchem jednostajnym z szybkością 45 m/s.

murzikaleks [220]

Lett me come back imma translate this... and then ill come to help

7 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