PHYSICS COLLEGE

We had a homework problem in which the Arrhenius equation was applied to the blinking of fireflies. Several other natural phenomena also obey that equation, including the temperature dependent chirping of crickets. A particular species, the snowy tree cricket, has been widely studied. These crickets chirp at a rate of 178 times per minute at 25.0°C, and the activation energy for the chirping process is 53.9 kJ/mol. What is the temperature if the crickets chirp at a rate of 126 times per minute?

Answers

Answer 1

Answer:

Answer:

Temperature = 20.35°C

Explanation:

Arrhenius equation is as follows:

k = A*exp(-Ea/(R*T)), where

k = rate of chirps

Ea = Activation Energy

R = Universal Gas Constant

T = Temperature (in Kelvin)

A = Constant

Given Data

Ea = 53.9*10^3 J/mol

R = 8.3145 J/(mol.K)

T = 273.15 + 25 K

k = 178 chirps per minutes

Calculation

Using the Arrhenius equation, we can find A,

A= 4.935x10^11

Now we can apply the same equation with the data below to find T at k=126,

k = A*exp(-Ea/(R*T))

Ea = 53.9*10^3

R = 8.3145

k = 126

T = 20.35°C

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Answers

Answer: 38.5rad/s

Explanation: The calculations can be viewed on the image attached below. Thanks

5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)

Answers

Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d

Where,

F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

$W = F*d\nW = mg*d\nW=281.5*9.8(17.1*10^(-2)\nW = 471.738 J\approx 472J$

PART B) Using Newton's Second law we have that,

$F = mg \nF= 281.5*9.8\nF= 2758.7 N \approx 2.76kN$

Give some reasons for our knowledge of the solar system has increased considerably in the past few years. Support your response with at least 3 reasons with details regarding concepts from the units learned in this course.

Answers

Answer:

Improvement in observational, and exploratory technology

Rapid increase in knowledge

International collaboration

Explanation:

Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.

Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.

Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.

International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Answers

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

W = $F_(total) .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) .d =(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) = ((1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i))/(d)$

$F_(total=) ((1)/(2) X 62 X6^(2) -(1)/(2) X 62 X2^(2) )/(25)$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_(sprinter) = F_(total) + F_(wind) = 39.7 + 30 = 69.68 N$

Answer:

Force exerted by sprinter = 69.68 N

Explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

Two cars, one with mass mi = 1200 kg is traveling north, and the other a large car with mass m2= 3000 kg is traveling East. If they collide while each car is traveling with a speed v = 5.0 m/s, what is the car's final speed and direction (vector notation can be given as well). Assume the collision is perfectly inelastic.

Answers

Answer:

Both cars travel at < 10 , 4 > m/s

Explanation:

Conservation of Linear Momentum

The total momentum of a system of particles of masses m1 and m2 traveling at velocities v1 and v2 (vectors) is given by

$\vec p_1=m_1\vec v_1+m_2\vec v_2$

When the particles collide, their velocities change to v1' and v2' while their masses remain unaltered. The total momentum in the final condition is

$\vec p_2=m_1\vec v'_1+m_2\vec v'_2$

We know the collision is perfectly inelastic, which means both cars stick together at a common final velocity v'. Thus

$\vec p_2=m_1\vec v'+m_2\vec v'=(m_1+m_2)\vec v'$

Both total momentums are equal:

$m_1\vec v_1+m_2\vec v_2=(m_1+m_2)\vec v'$

Solving for v'

$\displaystyle v'=(m_1\vec v_1+m_2\vec v_2)/(m_1+m_2)$

The data obtained from the question is

$m_1=1200\ kg$

$m_2=3000\ kg$

The first car travels north which means its velocity has only y-component

$\vec v_1=<0,5>$

The second car travels east, only x-component of the velocity is present

$\vec v_2=<5,0>$

Plugging in the values

$\displaystyle v'=(1200<0,5>+3000<5,0>)/(1200+3000)$

$\displaystyle v'=(<0,6000>+<15000,0>)/(1500)$

$\displaystyle v'=(<15000,6000>)/(1500)=<10,4>\ m/s$

The magnitude of the velocity is

$|v'|=√(10^2+4^2)=10.77\ m/s$

And the angle

$\displaystyle tan\alpha=(4)/(10)=0.4$

$\alpha=21.8^o$

A commercial aircraft is flying westbound east of the Sierra Nevada Mountains in California. The pilot observes billow clouds near the same altitude as the aircraft to the south, and immediately turns on the "fasten seat belt" sign. Explain why the aircraft experiences an abrupt loss of 500 meters of altitude a short time later.

Answers

Answer:

Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.

Explanation:

Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.

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