How is the air found in soils different from the air in the atmosphere?

Answers

Answer 1

Answer:

Air found in the soil is trapped in it and is not exposed to the air currents, poor  in oxygen and rich in moisture.

Explanation:

Air found in the soil is trapped in it and is not exposed to the air currents.

It contains more moisture than the atmospheric air.

It is rich in carbon dioxide and poor in oxygen.

The air found in the soil forms a part of the lithosphere and not the atmosphere.

Nitrogen trapped in the soil is used by the plants to make protein, carbon dioxide is used for the photosynthesis and oxygen is used for the respiration of roots and microorganisms.


Related Questions

The filament in an incandescent light bulb is made from tungsten. The resistivity of tungsten = 5.6e-8 Ω*m. The radius of the tungsten wire is 0.045 mm. If the bulb is plugged into a 120 V outlet and is to draw a current of 1.24 A, how long must the wire be?

Answers

Answer:

11m

Explanation:

Given:

Resistivity ρ = 5.6e-8 Ωm

Radius r = 0.045 mm [tex]=\frac{0.045}{1000}[/tex] = 4.5 x 10⁻⁵ m

Voltage V = 120V

Current I = 1.24A

From Ohm's law, [tex]R = \frac{V}{I}[/tex]

                            [tex]R = \frac{120}{1.24}[/tex]

                            R = 96.77 Ω

Resistivity = (Resistance × Area)/ length

      ρ = (RA)/L

Therefore, the length of a wire is given by;

       L = (RA)/ρ

Calculating the area A of the wire;

       A = πr²

       A = π × (4.5 x 10⁻⁵)²

       A = 6.36 x 10⁻⁹ m²

Substituting area of the wire A =  6.36 x 10⁻⁹ m² into the equation of the length of wire

       L = (96.77 × 6.36×10⁻⁹ ) / 5.6×10⁻⁸

       L = 10.9977m

       L = 11m (approximately)

Listed following are several objects in the solar system. Rank these objects from left to right based on the distance from the Sun at which they are presumed to have formed, from nearest to farthest.

Answers

a typical asteroid in the asteroid belt, a typical oort cloud object, a typical kuiper belt object

Although Oort cloud comets are now located far beyond the Kuiper belt, they are thought to have formed in the region of the jovian planets. They were then "kicked out" to their current orbits by gravitational encounters with the jovian planets.

Final answer:

The objects in the solar system can be ranked from nearest to farthest based on their formation distance from the Sun: Mercury, Venus, Earth, Mars, and Jupiter.

Explanation:

The objects in the solar system can be ranked based on their distance from the Sun as follows:

Mercury: This planet is the closest to the Sun and is presumed to have formed at its current location.Venus: Venus is the second planet from the Sun and also formed relatively close to it.Earth: Our home planet is the third planet from the Sun and formed within the habitable zone.Mars: Mars is the fourth planet from the Sun and formed farther away than Earth.Jupiter: As the largest planet in the solar system, Jupiter formed even farther away from the Sun than Mars.

Therefore, the ranking from nearest to farthest would be Mercury, Venus, Earth, Mars, and Jupiter.

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A 58.1g sample of quartz is put into a calorimeter that contains 250.0g of water. The quartz sample starts off at 98.4°C and the temperature of the water starts off at 25.0°C. When the temperature of the water stops changing it's 27.9°C. The pressure remains constant at 1 atm. Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to the correct number of significant digits.

Answers

Answer:

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

Explanation:

Given:

mass of sample quartz, [tex]m_q=58.1\ g[/tex]mass of water in calorie-meter, [tex]m_w=250\ g[/tex]initial temperature of quartz, [tex]T_{qi}=98.4^{\circ}C[/tex]initial temperature of water, [tex]T_{wi}=25^{\circ}C[/tex]final temperature of the mixture, [tex]T_{f}=27.9^{\circ}C[/tex]

We have:

specific heat capacity of water, [tex]m_c=4.186\ J.g^{-1}.^{\circ}C^{-1}[/tex]

Assuming that heat loss is neither from any of the components before mixing nor from the mixture just after mixing. Also the container does not absorbs any heat.

Therefore,

heat gained by the water = heat lost by the quartz

[tex]Q_w=Q_q[/tex]

[tex]m_w.c_w. (T_f-T_{wi})=m_q.c_q.(T_{qi}-T_f)[/tex]

    where: [tex]c_q=[/tex] specific heat capacity of quartz

[tex]250\times 4.186\times (27.9-25)=58.1\times c_q\times (98.4-27.9)[/tex]

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

Final answer:

To calculate the specific heat capacity of quartz, use the equation q = mcΔT and set it equal to the equation for the heat transferred to the water. Solve for c to find the specific heat capacity of quartz.

Explanation:

The specific heat capacity of quartz can be calculated using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat transferred to the water can be calculated using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity of water (4.184 J/g °C), and ΔT is the change in temperature.

By setting the two equations equal to each other and solving for c, the specific heat capacity of quartz can be determined.

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Tech A says the first external component to install when assembling an engine is the intake manifold. Tech B says the exhaust manifolds should be installed after the intake manifold. Who is correct?

Answers

Answer:

both technicians are correct

Explanation:

both technicians are correct because technician A says the first external component to be installed is the intake manifold and technician B says the exhaust manifold should be installed after the intake manifold has been installed, this implies the same thing because installing the exhaust manifold after installing the intake manifold means you are installing the intake manifold first.

Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform angular acceleration of 0.01 rad/s2 about the center O. In the figure, the magnitude of the linear acceleration of P, when it reaches the y-axis, is closest to:
a..063
b..075
c..072
d..069
e..066

Answers

Answer:

e). [tex]a = 0.066 m/s^2[/tex]

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta[/tex]

now we have

[tex]\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})[/tex]

[tex]\omega = 0.177 rad/s[/tex]

now the centripetal acceleration of the point P is given as

[tex]a_c = \omega^2 R[/tex]

[tex]a_c = (0.177)^2(2)[/tex]

[tex]a_c = 0.063 m/s^2[/tex]

tangential acceleration is given as

[tex]a_t = R\alpha[/tex]

[tex]a_t = 2(0.01)[/tex]

[tex]a_t = 0.02 m/s^2[/tex]

now net acceleration is given as

[tex]a = \sqrt{a_t^2 + a_c^2}[/tex]

[tex]a = \sqrt{0.02^2 + 0.063^2}[/tex]

[tex]a = 0.066 m/s^2[/tex]

Final answer:

The question involves the calculation of linear acceleration of a point on a wheel undergoing angular acceleration. The linear acceleration involves both tangential and radial components, and these are combined to provide the total acceleration. The closest answer is 0.066 m/s^2.

Explanation:

The subject of this question is physics, specifically the concepts of angular acceleration and linear acceleration. Given an angular acceleration, we can find the linear acceleration by using the formula a = rα, where a is linear acceleration, r is the radius, and α is angular acceleration. Substituting the given values, we get a = 2.0 m * 0.01 rad/s2 = 0.02 m/s2. This value is the tangential acceleration of point P.

Since the point P is on the y-axis, the linear acceleration, a, is given by the radial acceleration which is equal to ω2r, where ω is the angular velocity and r is the radius of the circle. But we also have ω2 = 2αr, where ρ is the angular acceleration and r is the radius. So, the radial acceleration is (2αr)2r = 4α2r3. Substituting for α and r we get: 4*(0.01 rad/s2)2*(2.0 m)3 = 0.0016 m/s2.

The total linear acceleration is the vector sum of the radial and tangential accelerations. Since these vectors are perpendicular, we calculate their resultant by Pythagoras' theorem: √(at2 + ar2) = √(0.022 m/s2 + 0.00162 m/s2) = 0.02016 m/s2. Hence, the choice that's closest is (a) 0.066 m/s2.

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What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then released from rest, and whose period is 0.66 s?
Assume that the displacement at the start of the motion is positive.
What will be its displacement after 2.3 s ?

Answers

Answer:

[tex]0.836cm[/tex] in the positive direction.

Explanation:

The equation that describes the motion of this mass-spring system is given by;

[tex]y=Asin\omega t...................(1)[/tex]

Where A is the amplitude, which defined as the maximum displacement from the equilibrium position for a body in simple harmonic motion.

[tex]\omega[/tex] is the angular velocity measured in [tex]rads^{-1[/tex], this is the angle turned through per unit time.

[tex]y[/tex] is the displacement along the axis of the amplitude, and [tex]t[/tex] is any instant of time in the motion.

Given; A = 8.8cm = 0.088m

The angular velocity is given by the following relationship also;

[tex]\omega=2\pi/T[/tex]

Where T is the period, which is defined as the time taken for a body in simple harmonic motion to make one complete oscillation.

Given; T=0.66s

Therefore;

[tex]\omega=2\pi/0.66\\\omega=3.03\pi rads^{-1[/tex]

Substituting all values into equation (1) we obtain the following;

[tex]y=0.088sin3.03\pi t................(2)[/tex]

Equation (2) is the equation describing the motion of the mass on the spring.

At an instant of time t = 2.3s, the displacement is therefore given as follows;

[tex]y=0.088sin[3.03\pi(2.3)]\\y=0.088sin6.97\pi\\[/tex]

By conversion, [tex]6.97\pi rad=6.97*180^o=1254.55^o[/tex]

Therefore

[tex]y=0.088sin1254.54=0.00836m\\[/tex]

[tex]y=0.00836m=0.836cm[/tex]

An engine does 19900 J of work and rejects 5300 J of heat into a cold reservoir whose temperature is 285 K. What would be the smallest possible temperature of the hot reservoir?

Answers

Answer:

[tex]T_H = 1335\ J[/tex]

Explanation:

Given,

Work done by engine = 19900 J

rejected energy of heat =  5300 J

temperature = 285 K

efficiency of the engine

[tex]\eta = \dfrac{W}{Q_H}[/tex]

where

[tex]W = Q_H - Q_C[/tex]

[tex]19900 = Q_H - 5300[/tex]

[tex]Q_H = 25200\ J[/tex]

efficiency in terms of temperature

[tex]\eta = \dfrac{T_H-T_C}{T_H}[/tex]

from the above two equation

[tex]\dfrac{Q_H-Q_C}{Q_H}= \dfrac{T_H-T_C}{T_H}[/tex]

[tex]1-\dfrac{Q_C}{Q_H}= 1- \dfrac{T_C}{T_H}[/tex]

[tex]\dfrac{Q_C}{Q_H}= \dfrac{T_C}{T_H}[/tex]

[tex]T_H=T_C\ \dfrac{Q_H}{Q_C}[/tex]

[tex]T_H=285\times \dfrac{25200}{5300}[/tex]

[tex]T_H = 1335\ J[/tex]

Final answer:

The smallest possible temperature of the hot reservoir, given the work done by the engine, heat rejected into the cold reservoir, and temperature of the cold reservoir, is approximately 1357.14 Kelvin.

Explanation:

The problem is based on the Carnot heat engine, a theoretical model used for understanding the thermodynamics of heat conversion into work. The efficiency of Carnot's engine depends on the temperatures of the hot and cold reservoirs (Th and Tc respectively). The efficiency, represented as e, is calculated as e = (Th-Tc) / Th.

Before we do the calculation, it's important to note that temperatures must be expressed in Kelvin for this formula to be accurate. Tc, the temperature of the cold reservoir, is given as 285 K. The work done by the engine, represented as W, is 19900 J, and the heat rejected by the engine into the cold reservoir, represented as Qc, is 5300J. The heat absorbed by the engine from the hot reservoir, represented as Qh, can be calculated using the first law of thermodynamics, which states Qh = W + Qc. Therefore, Qh = 19900 J + 5300 J = 25200 J.

Now, to find Th, we rearrange the efficiency formula to Th = Tc / (1 - e), where e is now W / Qh = 19900 J / 25200 J ≈ 0.79. Substituting Tc = 285 K and e ≈ 0.79 into the formula, Th = 285 K / (1 - 0.79) ≈ 1357.14 K. This is the smallest possible temperature of the hot reservoir for the given conditions.

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Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m2. The father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 N · m.(a) Calculate the rotational kinetic energy (in J) in the merry-go-round plus child when they have an angular velocity of 29.2 rpm. J(b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. revolutions(c) Again, using energy considerations, calculate the force (in N) the father must exert to stop the merry-go-round in seven revolutions. N

Answers

Final answer:

In this part, we discuss the expected decrease in angular acceleration when the child is on the merry-go-round due to the greater moment of inertia. We calculate the child's moment of inertia by approximating the child as a point mass at a specific distance from the axis.

Explanation:

b. We expect the angular acceleration for the system to be less in this part because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I, we first find the child's moment of inertia Ic by approximating the child as a point mass at a distance of 1.25 m from the axis. Then

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Final answer:

The rotational kinetic energy of the merry-go-round plus child at 29.2 rpm is 394.5834 J. The father must push for approximately 0.167 revolutions to reach this velocity from rest. A force of -8.975 N (in the opposite direction of motion) must be exerted to stop the merry-go-round in seven revolutions.

Explanation:

The problem addresses the concepts of angular velocity, rotational kinetic energy, and the application of work-energy principles within a physics context, specifically relating to a merry-go-round system.

(a) To calculate the rotational kinetic energy (KErot) of the merry-go-round plus child, we use the formula KErot = (1/2)Iω^2 where I is the moment of inertia and ω is the angular velocity in radians per second. First, convert the angular velocity from rpm to rad/s using the conversion factor 1 rpm = (2π/60) rad/s.
KErot = (1/2)(84.4 kg · m^2)(29.2 rpm × 2π/60 rad/s)^2
= (1/2)(84.4)(3.058 rad/s)^2
= (1/2)(84.4)(9.351)
= (1/2)(789.1668) J
= 394.5834 J

(b) The number of revolutions needed to achieve this angular velocity from rest can be found using the work-energy principle, where work done (W) by the father's force equals the rotational kinetic energy of the merry-go-round system.
W = τr(2πN)
394.5834 J = 375 N · m × (2πN)
N = 394.5834 J / (375 N · m × 2π)
N = 0.167 revolutions

(c) The force to stop the merry-go-round in seven revolutions can be calculated using the concept that the work done by the father's force (W) to stop the merry-go-round equals the negative change in the rotational kinetic energy. Thus, W = -KErot, and the rotational kinetic energy is initially computed in part (a). This gives:
W = -394.5834 J
Force (F) = W / (2π×7)
F = -394.5834 J / (2π × 7)
F = -8.975 N (negative sign indicates that the force is applied in the opposite direction of motion)

A type of cuckoo clock keeps time by having a mass bouncing on a spring, usually something cute like a cherub in a chair. What force constant (in N/m) is needed to produce a period of 0.660 s for a 0.0100 kg mass?

Answers

Answer:0.906 N/m

Explanation:

Given

time period [tex]T=0.66 s[/tex]

mass [tex]m=0.01 kg[/tex]

System can be considered as spring mass system

Time Period of spring mass system is given by

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

squaring

[tex]k=\frac{4\pi ^2m}{T^2}[/tex]

[tex]k=\frac{4\times \pi ^2\times 0.01}{0.66^2}[/tex]

[tex]k=0.906 N/m[/tex]

What can turn a natural hazard into a natural disaster?

Answers

Answer: A hazard is a condition that has the potential to cause harm.

A natural hazard is a potentially harmful situation, where a person places himself in a naturally unsafe zone.

A natural disaster is a large scale destruction of life and properties by the forces of nature.

A natural hazard can become a natural disaster in some cases where the natural harmful situation a person places himself in, is acted upon by the forces of nature on a large scale.

Based on discoveries to date, which of the following conclusions is justified?a) Most stars have one or more terrestrial planets orbiting within their habitable zones.b) Planets are common, but planets as small as Earth are extremely rare.c) Planetary systems are common and planets similar in size to Earth are also common.d) Although planetary systems are common, few resemble ours with terrestrial planets near the Sun and jovian planets far from the Sun.

Answers

Based on discoveries to date, the conclusion as “Planetary systems are common and planets similar in size to Earth are also common” is justified.

Answer: Option C

Explanation:

Some studies show that on average, each star has at least single planet. This means that most stars, such as the Solar System, possess planets (otherwise exoplanets). It is known that small planets (more or less Earthly or slightly larger) are more common than giant planets. The mediocrity principles state that planet like Earth should be universal in the universe, while the rare earth hypothesis says they are extremely rare.

Size is often considered an important factor, because planets the size of the Earth are probably more terrestrial and can hold the earth's atmosphere. The planetary system is a series of gravitational celestial objects orbiting a star or galaxy. Generally, planetary systems describe systems with one or more planets, although such systems may also consist of bodies such as dwarf planets, asteroids and the like.

To calculate current flow through any branch of a circuit by substituting the values of IX and RX for the branch
values when total circuit current and the resistance are known, use the _______________ formula.

A. reciprocal
B. current divider

Answers

Answer:

Current divider

Explanation:

To calculate current flow through any branch of a circuit by substituting the values of [tex]I_X\ and\ R_X[/tex] for the branch  values when total circuit current and the resistance are known, use the current divider formula.

A current divider is a circuit that produces output current as a function of input current. It is a rule to find the splitting of the current in all branches of the circuit. Hence, the correct option is (B).  

Final answer:

The formula to calculate the current flow through any branch of a circuit when the total circuit current and resistance are known is the current divider formula.

Explanation:

To calculate the current flow through any branch of a circuit by substituting the values of IX and RX for the branch values when total circuit current and the resistance are known, use the current divider formula. The current divider rule is particularly useful in parallel resistor circuits and it allows for the easy calculation of the current flowing through a resistor in parallel.

In a parallel circuit, the voltage across each branch is the same, but the current through each branch can be different, depending on the resistance of that branch. The current divider rule states that the current through a branch is the ratio of the total parallel resistance to the branch resistance, times the total current entering the parallel combination.

For example, if we want the current through resistor R1 (I1), and we know the total current (Itot) and the total parallel resistance (Rp), as well as the resistance of R1 (R1), we can use the formula:

I1 = Itot * (Rp / R1)

By knowing the total resistance and the total current, one can deduce the individual branch currents using Ohm's Law and the principles of parallel circuits.

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Most of the water vapor and carbon dioxide in earth's atmosphere is found A. In the upper part of the atmosphere. B.In the thin layer right in the middle of the atmosphere C.In the lower part of the atmosphere. D.In the thick layer right in the middle of the atmosphere

Answers

Answer:

C] In the lower part of atmosphere

As the gases ( O₂ & CO₂ ) are most abundant here and supports life.

Hope it helps!

Happy Learning!!

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A freight car with a mass of 10 metric tons is rolling at 108 km/h along a level track when it collides with another freight car, which is initially at rest. If the speed of the cars after they couple together is 36 km/h, what is the mass of the second car?

Answers

Answer:

20 metric tons

Explanation:

Using the Principle of Conservation of Linear Momentum which states that the total momentum of a system is constant in any direction in which no external force acts.

momentum before collision= momentum after collision

mathematically expressed as

m1u1 + m2u2 = m1v1+m2v2 =0

where:

m1= mass of rolling freight car = 10 metric tons = 10Mg

1 ton = 1000kg= 1Mg

u1 =initial velocity of freight car = 108km/h,

m2 = mass of stationary freight car = ?, this the unknown we are finding

u2 = initial velocity of

m2 = 0, because it is at rest,

v1= final velocity of m1,

v2= final velocity of m2

since they move together in the same direction they share a common final velocity as such

v1=v2= 36km/h

substituting valves in expression give

m1u1 + m2u2 = m1v1 + m2v2= 0

10Mgx 108km/hr + m2 x 0 = 10Mg x 36km/h + m2x36km/h

1080( Mg x km/hr) + 0 = 360 (Mg x Km/h) + 36m2 (km/h)

1080 (Mgx km/hr) - 360(Mg x km/h) = 36m2 (km/h)

720(Mg x km/h) = 36m2 (km/h)

divide both side by  36 (km/h) gives

m2 = 20Mg =20 metric tons

Answer: B). 20 metric tons

Explanation: If a freight car with a mass of 10 metric tons is going at 108 km/h collides with another freight car, that is parked. If the speed of the cars, after they crash together, is 36 km/h, the mass of the second car is 20 metric tons.

I AM 100% CONFIDENT IN THIS ANSWER!

A new gel is being developed to use inside padding and helmets to cushion the body from impacts. The gel is stored in a 4.1 m^3 cylindrical tank with a diameter of 2 m. The tank is pressurized to 1.3 atm of surface pressure to prevent evaporation. A total pressure probe located at the bottom of the tank reads 60 ft of water. What is the specific gravity of the gel contained in the tank?

Answers

Answer:

[tex]SE_{gel} = 3.75[/tex]

Explanation:

First, we have to calculate the gel's column height using the cylinder's volume, as follows:

[tex]V=\pi\times r \times h\\h=\frac{V}{\pi \times r}\\h=\frac{4.1m^3}{\pi \times 1m}= 1.30 m[/tex]

Then, as the pressure given at the bottom of the tank is the sum of the surface pressure and the gel's column pressure, we need to calculate only the gel's column pressure:

ft of water is a unit of pressure, but we need to convert it to atm and then to Pa, in order to calculate our results in the correct units. Therefore, the conversion factor is:

1 ft of water (4°C) = 0.0295 atm

[tex]60 ft water \times \frac{0.0295 atm}{1 ft water}= 1.77  atm\\P_{bottom}=P_{surface}+P_{gel}\\P_{gel}=P_{bottom}-P_{surface}=1.77 atm - 1.3 atm\\P_{gel}= 0.47 atm\times \frac{101325Pa}{1 atm}=47622.75 Pa[/tex]

Now, to calculate the specific gravity, we need to find first the gel's density:

[tex]P_{gel} = \rho gh\\\rho = \frac{P_{gel}}{gh}=\frac{47622.75 Pa}{9.8 m/s^2 \times 1.30m}= 3738.04 \frac{kg}{m^3}[/tex]

[tex]SE_{gel} = \frac{\rho_{gel}}{\rho_{water}}= \frac{3738.04 kg/m^3}{997 kg/m^3} = 3.75\\SE_{gel} =3.75[/tex]

The specific gravity of the gel is 3.75.

Final answer:

To calculate the specific gravity of a gel, convert the pressure reading from feet of water to pascals, add the tank's pressurized atmosphere, then use the relation between pressure, density, gravity, and height to solve for the density of the gel, which is then compared to the density of water.

Explanation:

The question is asking to calculate the specific gravity of a gel based on the pressure reading from a probe at the bottom of a cylindrical tank. We are given the pressure as 60 ft of water. To convert this to a pressure that we can use to find density, we need to convert feet of water to pascals:

1 ft H2O = 2989.07 Pa

60 ft H2O = 60 * 2989.07 Pa = 179344.2 Pa

Since the tank is pressurized to 1.3 atmospheres at the surface, we must also consider this in our pressure calculation:

1 atm = 101325 Pa

1.3 atm = 1.3 * 101325 Pa = 131722.5 Pa

The total pressure at the bottom of the tank is the sum of the pressure due to the gel and the pressurized air:

Total pressure = 179344.2 Pa + 131722.5 Pa

Total pressure = 311066.7 Pa

To find the specific gravity, we use the following relation where the specific gravity is the ratio of the density of the gel (ρgel) to the density of water (ρH2O):

Specific gravity = ρgel / ρH2O

We know that pressure is also the product of density (ρ), gravity (g=9.81 m/s²), and height (h=60 ft * 0.3048 m/ft) for the fluid:

Pressure = ρgel * g * h

ρgel = Pressure / (g * h)

ρgel = 311066.7 Pa / (9.81 m/s² * 60 ft * 0.3048 m/ft)

After calculating ρgel, we divide that by the density of water (1000 kg/m³) to find the specific gravity.

Amy is out with her friends. Misfortune occurs and Amy and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel station. Determine the work done on the car.A. 2.35 x 105 JB. 2.05 x 105 JC. 1.35 x 105 JD. 3.25 x 105 J

Answers

Answer:

2.35 x 10^{5} J

Explanation:

force (f) = 1080 N

distance (d) = 218 m

find the work done

work is said to be done when force applied to an object moves the object, therefore work done = force x distance

work done = 1080 x 218 = 235,440 J = 2.35 x 10^{5} J

A uniform meter stick of mass $M$ has a half-filled can of fruit juice of mass $m$ attached to one end. The meter stick and the can balance at a point 21.2 cm from the end of the stick where the can is attached. When the balanced stick-can system is suspended from a scale, the reading on the scale is 2.58 N. Calculate the mass of the meter stick.

Answers

Answer:

111.6 g

Explanation:

Given

m + M = 2.58 / 9.8

= 0.2632 kg

When the can of fruit juice is balance from scale , we get following relation

M x ( 50 - 21.2 ) = m x 21.2 ( balancing the torque due to weight of scale and can about the balancing point )

M x 28.8 = 21.2 m

= 21.2 ( 0.2632 - M )

= 5.58 - 21.2 M

M ( 28.8 + 21.2 ) = 5.58

M = .1116 kg

= 111.6 g

An object is acted upon by a constant force which gives it a constant acceleration a. At a certain time t1, having started from rest at t = 0, it has kinetic energy K1. At what time t2 has its kinetic energy doubled?

Answers

Answer:

[tex]t_2=\sqrt{2}(t_1)[/tex]

Explanation:

The kinetic energy of a body is that energy it possesses due to its motion. In classical mechanics, this energy depends only on its mass and speed, as follows:

[tex]K=\frac{mv^2}{2}[/tex]

The speed in an uniformly accelerated motion is given by:

[tex]v=at[/tex]

Replacing this expression in the formula for the kinetic energy, we have:

[tex]K=\frac{ma^2t^2}{2}\\[/tex]

So, if we have [tex]K_2=2K_1[/tex]:

[tex]K_1=\frac{ma^2t_1^2}{2}(1)\\K_2=\frac{ma^2t_2^2}{2}\\2K_1=\frac{ma^2t_2^2}{2}\\K_1=\frac{ma^2t_2^2}{4}(2)\\[/tex]

Equaling (1) and (2) and solving for [tex]t_2[/tex]:

[tex]\frac{ma^2t_1^2}{2}=\frac{ma^2t_2^2}{4}\\t_2=\frac{4t_1^2}{2}\\t_2=\sqrt{2t_1^2}\\t_2=\sqrt{2}(t_1)[/tex]

The Earth and the moon are attracted to each other by gravitational force. The more massive Earth attract the less massive moon with a force that is (greater than, less than, the same as) the force with which the moon attracts the Earth.

Answers

Answer:

Earth attract the less massive moon with a force that is the same as the force with which the moon attracts the Earth.

Explanation:

This can be explained by Newton's third law:

[tex]F_{12}=-F_{21}[/tex]

The force exerted by body 1 on body 2 is the same as that exerted by body 2 on body 1, only with the opposite sign.

In this case that force is the gravitational force, but the law still applies.

So the moon and the earth are attracted with the same magnitude of force.

In the experiment, you will determine the wavelength of a laser by shining the laser beam on a diffraction grating, a set of regularly spaced lines. If the screen is 10 m away, the spots are 5.4 cm apart, and the lines of the grating are separated by 0.1 mm,
what is (in nm)?

Answers

Answer:

The wavelength is 539 nm.

Explanation:

Given that,

Distance of screen L= 10 m

Width s= 5.4 cm

Grating separation d= 0.1 mm

Suppose the pattern is displayed on a screen a distance L from the grating and the spots are separated by s. we need to find the wavelength.

The angle between the central and first maximum is given as

[tex]\tan\theta=\dfrac{s}{L}[/tex]

[tex]\theta=\tan^{-1}\dfrac{s}{L}[/tex]

[tex]\theta=\tan^{-1}\dfrac{5.4\times10^{-2}}{10}[/tex]

[tex]\theta=0.309[/tex]

We need to calculate the wavelength

The condition for maximum of a diffraction grating is

[tex]d\sin\theta=m\lambda[/tex]

[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{0.1\times10^{-3}\times\sin0.309}{1}[/tex]

[tex]\lambda=539\times10^{-9}\ m[/tex]

[tex]\lambda=539\ nm[/tex]

Hence, The wavelength is 539 nm.

Final answer:

The wavelength of a laser can be determined using the diffraction grating equation. Calculate the θ value with θ = tan⁻¹(y/L), then substitute the known values into the equation λ = (d sin θ) / m. Convert the resultant wavelength to nanometers.

Explanation:

Your question relates to the physics of light and diffraction grating. We can find the wavelength of the laser by using the diffraction grating equation. The equation is: d sin θ = mλ

Where:

d is the grating spacing (distance between the lines represented in meters)θ is the angle to the m'th order maximumm is the order of maximum (in your case, it's likely 1)λ is the wavelength we are trying to find

In your case, we need to calculate θ first: θ = tan⁻¹(y/L) where y is the distance between the spots (5.4cm) and L is the distance from the grating to the screen (10m).

Substitute the known values and solve for λ: λ = (d sin θ) / m

Remember, the resultant wavelength should be converted to nanometers (1m = 1 x 10⁹ nm).

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Two balls are thrown horizontally. Ball C is thrown with a force of 20 N, and ball D is thrown with a force of 40 N. Assuming all other factors are equal, ball D will fall toward the ground

Answers

Answer:

ball D will fall toward the ground at the same time as ball C

Explanation:

both balls experience the same downward (vertical) force of gravity as such they will both fall down at the same time, given that all other factors are equal.

although the ball were through with different forces,

those forces where in the horizontal direction but the force of gravity (downward force) will act on them equally to bring them down at the same time

Termination of translation requires a termination signal, RNA polymerase, and a release factor. a release factor, initiator tRNA, and ribosomes. initiation factors, the small subunit of the ribosome, and mRNA. elongation factors and charged tRNAs.

Answers

termination signala release factor

Termination of translation requires a termination signal, and a release factor

Explanation:

There are 3 stops codons of the 64 possible codons. These are UAA, UAG, or UGA. These do not code for amino acids and are therefore not recognized by any anticodons for any of the ‘charged’ tRNA. These codons, are recognized by release factors that ‘knock off’ the newly synthesized peptide from the ribosome through peptidyl-tRNA hydrolysis. There are several release factors (RF1 and RF2) in bacteria but in eukaryotes only one RF has been discovered to recognize the 3 stop codons.

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Final answer:

Translation termination is signaled by a stop codon and facilitated by release factors and ribosomes, which lead to the release of the newly synthesized protein and the disassembly of the ribosomal complex.

Explanation:

Termination of translation requires a termination signal, which is a stop codon (UAA, UAG, or UGA) that doesn't code for an amino acid but rather signals the end of the translation process. During this stage, release factors recognize the stop codon and prompt the addition of a water molecule to the carboxyl end of the peptidyl-tRNA in the P site. This action leads to the release of the newly synthesized polypeptide chain. Ribosomes, which consist of a large and a small subunit, dissociate from the mRNA and from each other upon the completion of translation.

Ribosomes, release factors, and the mRNA template are the key participants in the termination process. Initiator tRNA and elongation factors with charged tRNAs are involved in the initiation and elongation stages of translation, but not in termination.

A heavy steel ball is hung from a cord to make a pendulum. The ball is pulled to the side so that the cord makes a 5 ∘ angle with the vertical. Holding the ball in place takes a force of 40 N . If the ball is pulled farther to the side so that the cord makes a 8 ∘ angle, what force is required to hold the ball?

Answers

Answer:

f = 63.8 N

Explanation:

initial angle to the vertical = 5 degrees

initial holding force = 40 N

final angle to the vertical = 8 degrees

final holding force = ?

find the final holding force

force = mgSinθ

        m = mass  and  g = acceleration due to gravity

for the initial holding force:

        40 = mgSin5

        mg = [tex]\frac{40}{sin5}[/tex] ....equation 1

for the final holding force:

        f = mgSin8  ......equation 2

   substituting the value of mg from equation 1 (where mg = [tex]\frac{40}{sin5}[/tex] ) into equation 2

    f = mgSin8 = [tex]\frac{40}{sin5}[/tex] x Sin8

   f = 63.8 N

       

Answer:

64 N

Explanation:

Given that the angles are very small, the following approximation can be made:

F ≈ m*g*α

where F is the force needed to hold the ball, m is the ball mass, g is the acceleration of gravity and α is the angle between the cord and the vertical.

Let's call F1 the force needed to hold the ball at 5° (α1) and F2 the force needed to hold the ball at 8° (α2).

F1 ≈ m*g*α1

F2 ≈ m*g*α2

Dividing the equations:

F1/F2 = α1/α2

F2 = (α2/α1)*F1

F2 = (8/5)*40

F2 = 64 N

A race car travels 99 m/s around a circular track of radius 52 m. What is the magnitude of the resultant force on the 1900 kg driver and his car if the car does not slip? Answer in units of kN.

Answers

Final answer:

The magnitude of the resultant force on the race car and driver is 357.375 kN.

Explanation:

First, we need to find the centripetal acceleration of the race car traveling around the circular track. The centripetal acceleration can be found using the formula:

a = v² / r

where v is the velocity of the car and r is the radius of the circular track. Plugging in the values, we get:

a = (99 m/s)² / 52 m = 188.25 m/s²

Next, we can use Newton's second law of motion to find the magnitude of the resultant force acting on the car. The formula is:

F = m * a

where F is the force, m is the mass of the car, and a is the centripetal acceleration. Plugging in the values, we get:

F = (1900 kg) * (188.25 m/s²) = 357,375 N

Finally, we can convert the force to kilonewtons (kN) by dividing by 1000:

F = 357,375 N / 1000 = 357.375 kN

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Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N. What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?

Answers

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

[tex]F=ma[/tex]

[tex]100=100m\times a[/tex]

[tex]ma=1\ N[/tex]

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

[tex]F_{100-99}=ma[/tex]

[tex]F_{100-99}=1[/tex]

We need to calculate the total number of masses attached to the string

Using formula for mass

[tex]m'=(100-50)m[/tex]

[tex]m'=50m[/tex]

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

[tex]F_{50}=m'a[/tex]

Put the value into the formula

[tex]F_{50}=50m\times a[/tex]

[tex]F_{50}=50\times1[/tex]

[tex]F_{50}=50\ N[/tex]

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

We have that for the Question "What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?"

it can be said that

The tension in the string connecting block 100 to block 99 = [tex]1N[/tex]The tension in the string connecting block 50 to block 51 = [tex]50N[/tex]

From the question we are told

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N.

Assuming mass of each block is 1 kg

The equation for the force is given as

[tex]F = ma\\\\a = \frac{F}{m}\\\\ = \frac{100}{100*1}\\\\ = 1 m/s^2[/tex]

Now, between block 100 and 99,

[tex]F = ma\\\\F = 1*1 \\\\= 1 N[/tex]

Now between block 50 and 51. There are 50 blocks behind 51 st block,  so,

[tex]m = 50 Kg\\\\a = 1 m/s2 (assuming all blocks accelerate at same rate)\\\\F = 50 * 1\\\\F= 50 N[/tex]

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A weather balloon is designed to expand to a maximum radius of 21 m at its working altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled at atmospheric pressure and 323 K, what is its radius at liftoff?

Answers

Answer:

7.65 m

Explanation:

[tex]P_1[/tex] = Initial pressure = 0.03 atm

[tex]P_2[/tex] = Final pressure = 1 atm

[tex]r_1[/tex] = Inital radius = 21 m

[tex]V_1[/tex] = Intial volume of gas = [tex]\frac{4}{3}\pi r_1^3[/tex]

[tex]V_2[/tex] = Final volume of gas = [tex]\frac{4}{3}\pi r_2^3[/tex]

[tex]T_1[/tex] = Initial temperature = 200 K

[tex]T_2[/tex] = Final temperature = 323 K

From ideal gas law we have

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m[/tex]

The radius at liftoff is 7.65 m

Final answer:

The initial radius of the weather balloon at liftoff can be calculated using the ideal gas law by comparing the conditions at the ground and at the balloon's working altitude.

Explanation:

The subject of this question deals with the physics principle of the behavior of gases, namely the ideal gas law. The ideal gas law links the pressure, temperature and volume of a gas, represented by the equation PV=nRT. Here, 'P' denotes pressure, 'V' is volume, 'n' is the number of moles, 'R' is the universal gas constant, and 'T' is temperature.

In the case of the weather balloon, when it is filled at atmospheric pressure and 323 K on the ground, and then expands to a maximum radius of 21m at a working altitude where the air pressure is 0.030 atm and the temperature is 200K, the principle of the ideal gas law can be applied. Assuming the same quantity of gas (n) we have P1V1/T1=P2V2/T2 where indices 1 and 2 refer to conditions at lift off and at altitude respectively.

We know that the volume of the balloon (V) is related to its radius (r) by the equation for volume of a sphere V=(4/3)*pi*r^3. So we can substitute the volumes in above equation with their expressions in terms of radius and solve for r1 (radius at liftoff).

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What is the magnitude φinitial of the magnetic flux through one turn of the coil before it is rotated?

Answers

Answer:

Explanation:

Facts about magnetic flux;

(1). Change in magnetic flux is equal to the current generated.

(2). The magnetic field moves from the north to the south.

(3). Magnetic flux is equal to how much magnetic field goes through a direction.

(4). Magnetlc direction can be found found by using Lenz's law.

Faraday's law= emf= N(BA/ ∆t.

Where N= number of loop

∆B/∆T = rate at which flux changes.

The magnitude φ(initial) of the magnetic flux through one turn of the coil before it is rotated is

φ(initial) = BA while φ(final)= zero(0).

The initial magnetic flux through one turn of the coil, with the coil initially perpendicular to Earth's magnetic field, is 6.10 × 10⁻⁹ T×m².

To calculate the magnitude of the initial magnetic flux (φinitial) through one turn of the coil, we use the formula:

φ = B × A × cos(θ)

where:

B is the magnetic field (5.00 × 10⁻⁵ T)A is the area the coil encloses (12.2 cm²)θ is the angle between the magnetic field and the normal to the plane of the coil

Since initially, the plane of the coil is perpendicular to Earth's magnetic field, θ = 0°, and hence cos(θ) = 1. First, we must convert the area from cm² to m²:

A = 12.2 cm² × (1 m² / 10,000 cm²) = 12.2 × 10⁻⁴ m²

Now, we can calculate the magnetic flux:

φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m² × cos(0°)

φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m²

φinitial = 6.10 × 10⁻⁹ T×m²

This is the magnitude of the initial magnetic flux through one turn of the coil before it is rotated.

Complete Question:

A coil has 230 turns enclosing an area of 12.2 cm³. In a physics laboratory experiment, the coil is rotated during the time interval 3.00 × 10⁻² s from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.00 × 10⁻⁵ T.

What is the magnitude φinitial of the magnetic flux through one turn of the coil before it is rotated?

(a) How far must the spring be compressed for 3.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.20 kg book onto it from a height of 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.

Answers

Answer:

a= 0.063 m

b = 0.116 m

Explanation:

First of all, we need the spring constant in order to solve this problem. You are not giving that data, but I will tell you how to solve this assuming a value of k, In this case, let's assume the value of k is 1600 N/m. (I solved an exercise like this before, using this value).

Now, we need to use the expressions to calculate the distance of the spring.

The elastic potential energy (Uel) is given with the following formula:

Uel = 1/2 kx²

Solving for x:

x = √2*Uel/k

Replacing the data in the above formula (And using the value of k os 1,600):

x = √2 * 3.2 / 1600

x = 0.063 m

b) For this part, we need to apply the work energy theorem which is:

K1 + Ugrav1 + Uel1 + Uo = K2 + Ugrav2 + Uel2

Since in this part, the exercise states that the book is dropped, we can say that the innitial and the end is 0, therefore, K1 = K2 = 0.

The spring at first is not compressed, so Uel1 = 0, and Uo which is the potential energy of other factors, is also 0, because there are no other force or factor here. Therefore, our theorem is resumed like this:

Ugrav1 = Uel2

The potential energy from gravity is given by:

Ug = mgy

And as the spring is placed vertically, we know the height which the book is dropped, so the distance y is:

y = x + h

And this value of x, is the one we need to solve. Replacing this in the theorem we have:

mg(h+x) = 1/2kx²

g would be 9.8 m/s²

Now, replacing the data:

1.2*9.8(0.8 + x) = 1/2*1600x²

Rearranging and solving for x we have:

1.2*9.8*2(0.8 + x) = 1600x²

18.82 + 23.52x = 1600x²

1600x² - 23.52x - 18.82 = 0

Now we need to solve for x, using the general formula:

x = - (-23.52) ± √(-23.52)² - 4 * 1600 * (-18.82) / 2*1600

x = 23.52 ± √553.19 + 120,448 / 3200

x = 23.52 ± 347.85 / 3200

x1 = 23.52 + 347.85 / 3200 = 0.116 m

x2 = 23.52 - 347.85 / 3200 = -0.101 m

Using the positive value, we have that the distance is 0.116 m.

A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light in the material?

Answers

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸  m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸  m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

Answer:

The percentage of speed of light in vacuum to the speed of light in the said material is 80%

Explanation:

The common values of refractive index are between 1 and 2 since nothing can travel faster than the speed of light, therefore, no material has a refractive index lower than 1.

According to the formula [tex]n=\frac{c}{v}[/tex]

where [tex]n[/tex] is the index of refraction

[tex]c[/tex] is the speed of light in vacuum

and [tex]v[/tex] is the speed of light in the material, it can be seen that n and v are inversely proportional which means greater the refractive index lower is the speed  of light.

Since we know that speed of light in vacuum is 300,000 km/s using the formula we get,

[tex]v=\frac{c}n}[/tex]

[tex]v=\frac{300000}{1.25}=240,000  km/s[/tex]

for finding percentage,

[tex]=\frac{240000}{300000}*100 = 80[/tex] %

The intensity of a sunspot is found to be 3 times smaller than the intensity emitted by the solar surface. What is the approximate temperature of this sunspot if the temperature of the solar surface is 5800 K?

Answers

Answer:

4400 K approximately

Explanation:

Stefan-Boltzmann law establish that the power radiated from a black body in terms of its temperature which is proportional to [tex]T^4[/tex]

[tex]i = aT^4[/tex]  with a Stefan-Boltzmann constant

Also we know that [tex]i_{sunspot}[/tex] is three times [tex]i_{sun}[/tex]

[tex]i_{sun}=3i_{sunspot}[/tex] or [tex]\frac{i_{sun}}{i_{sunspot}} = 3[/tex]

Using the Stefan-Boltzmann we can write

[tex]\frac{i_{sun}}{i_{sunspot}} = 3 = \frac{aT_{sun}^4}{aT_{sunspot}^4}[/tex]

solving for [tex]T_{sunspot}[/tex]

[tex]T_{sunspot} = \left(\frac{T_{sun}^4}{3}\right)^{1/4}[/tex]

Replacing the value of [tex]T_{sun}[/tex] (5800 K) it is obtained that [tex]T_{sunspot}[/tex] is 4407.05

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