The number density of an ideal gas at stp is called the loschmidt number. True or False

Answer:

B. OSMOTIC PRESSURE WILL BE LOWER IN THE ARTERIOLE END OF THE CAPPILLARY BED COMPARED TO THE VENOUS END.

Explanation:

This is true for filtration to take place in the cappillary bed. Osmotic pressure is the net pressure that drives movement of fluid from the interstitial fluid back into the capillaries. Osmotic pressure increase favors reabsorption as water moves from region of higher water concentration in the interstitial fluid to the lower region of water concentration in the capillaries.

At the ends of a capillary bed, the difference in the hydrostatic and osmotic pressures provides a net filtration or reabsorption ratio. At the arteriole end of the capillary bed, hydrostatic pressure is greater than the osmotic pressure allowing movements of fluid to the interstitial fluid (filtration) while as the blood moves to the venous end, the osmotic pressure becomes greater than than hydrostatic pressure.

Osmotic pressure is usually higher at the arteriole end of the capillary bed than at the venous end (Option C). This happens because plasma proteins remain in the capillary causing water to move back into the capillary.

In general, it is expected that osmotic pressure will be higher in the arteriole end of the capillary bed compared to the venous end (option C). This is because during capillary exchange, fluids and solutes are filtered out at the arteriole end of capillaries due to higher blood pressure, and then reabsorbed at the venous end due to higher osmotic pressure. This helps maintain fluid balance and prevent edema.

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Answer:

5174.4 J

Explanation:

Parameters given:

Mass of calorimeter, m = 1.4 kg = 1400 g

Specific heat capacity, c = 3.52 J/g°C

Temperature difference, ΔT = 28.5 - 27.45 = 1.05 °C

Heat absorbed by reaction, Q = m * c * ΔT

Q = 1400 * 3.52 * 1.05

Q = 5174.4 J

Answer:

5174.4Joules

Explanation:

Heat capacity is defined as the quantity of heat required to raise the temperature of total mass of a substance by 1Kelvin. Mathematically,

Q = mc∆t where;

Q is the amount of heat absorbed (in Joules)

m is the mass of the substance (bomb calorimeter) in g or kg

c is the specific heat capacity of the bomb calorimeter in J/g°C

∆t is the change in temperature in °C

Given m = 1.4kg

Since 1kg = 1000g

1.4kg = (1.4×1000)g

m = 1.4kg = 1400g

c = 3.52J/g°C

∆t = final temperature - initial temperature

Since heat was absorb (heat gained), final temperature will be 28.5°C

Initial temperature = 27.45°C

Substituting the data given into the heat capacity formula will give us;

Q = 1400×3.52×(28.5-27.45)

Q = 1400×3.52× 1.05

Q = 5174.4Joules

Amount of heat absorbed by the reaction is 5174.4Joules

Answer:

1600 cal

Explanation:

The formula

q=mΔ[tex]_{fus}[/tex][tex]H^{0}[/tex]

is used to calculate the heat required to melt a solid where q=amount of heat, m=mass and Δ[tex]_{fus}[/tex][tex]H^{0}[/tex] is the enthalphy of fusion

now we substitute, m=20g, Δ[tex]_{fus}[/tex][tex]H^{0}[/tex]=80cal/g

Therefore, q=20g x 80cal/g =1600 cal

I hope you find this information useful and interesting! Good luck!

The heat released when 20.0 g of water at 0°C is frozen to ice is 1600 calories.

When water freezes, it releases heat energy. The quantity of heat energy needed to convert a substance from a solid to a liquid or vice versa without affecting its temperature is known as the heat of fusion.

In this case, we are asked to find the number of calories of heat released when moment 20 g of water freezes to ice at zero degrees.

To calculate this, we can use the formula:

Heat released = mass of water * heat of fusion

Substituting the given values:

Heat released = 20.0 g * 80. cal/g = 1600 cal

Therefore, 1600 calories of heat are released when 20.0 g of water freezes to ice at 0 °C.

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Answer:

The answer to the question is;

1637.769 grams of water will need to be perspired in order to maintain his original temperature.

Explanation:

Energy intake of the person = 4000 kJ

Energy required to vaporize 1 mole of water = 44.0 kJ

That is 44.0 kJ/mole

Therefore

The number of moles of water that can be vaporized by 4000 kJ is given by

(4000 kJ)/ (44.0 kJ/mole) = 90.91 moles.

Mass of one mole of water = Molar mass of water = 18.01528 g/mol

Since number of moles of water = ([tex]\frac{Mass .of. water}{Molar. mass. of. water}[/tex])

We therefore have

Mass of water = (Number of moles of water)× (Molar mass of water)

Mass of water = 90.91 moles× 18.01528 g/mol = 1637.769 g

The mass (in grams) of water that he  would need to perspire in order to maintain his original temperature is 1637.769 g.

The person would need to perspire approximately [tex]\( 1638.86 \, \text{g} \)[/tex] (or about 1.64 kg) of water to maintain their original temperature after consuming 0.50 pounds of cheese with an energy intake of 4000 kJ.

To determine the mass of water the person needs to perspire to maintain their original temperature, we need to calculate the amount of water required to dissipate the energy intake through perspiration. Given:

Energy intake: [tex]\( 4000 \, \text{kJ} \)[/tex]

Heat of vaporization of water: [tex]\( 44.0 \, \text{kJ/mol} \)[/tex]

First, we need to find out how many moles of water need to be vaporized to dissipate 4000 kJ.

[tex]\[ \text{Moles of water} = \frac{\text{Total energy intake}}{\text{Heat of vaporization per mole}} = \frac{4000 \, \text{kJ}}{44.0 \, \text{kJ/mol}} \][/tex]

[tex]\[ \text{Moles of water} = \frac{4000}{44.0} \approx 90.91 \, \text{moles} \][/tex]

Next, we convert the number of moles to mass. The molar mass of water [tex](\( \text{H}_2\text{O} \))[/tex] is approximately [tex]\( 18.015 \, \text{g/mol} \).[/tex]

[tex]\[ \text{Mass of water} = \text{Moles of water} \times \text{Molar mass of water} = 90.91 \, \text{moles} \times 18.015 \, \text{g/mol} \][/tex]

[tex]\[ \text{Mass of water} = 1638.86 \, \text{g} \][/tex]

Answer:

The phase constant is 7.25 degree  

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = [tex]\sqrt{(\frac{v_x}{\omega })^2+y^2 }[/tex]  = [tex]\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }[/tex]  =  6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ  = 0.992

Ф = 7.25 degree  

so the phase constant is 7.25 degree  

Answer:

Explanation:

Given

Power output [tex]P=5\ kW[/tex]

efficiency [tex]\eta =25\ \%[/tex]

Engine expels [tex]Q_r=8\times 10^3\ J[/tex]

Efficiency is given by

[tex]\eta =1-\dfrac{Q_r}{Q_s}[/tex]

where [tex]Q_s[/tex]=Heat supplied

[tex]0.25=1-\dfrac{8\times 10^3}{Q_s}[/tex]

[tex]0.75=\dfrac{8\times 10^3}{Q_s}[/tex]

[tex]Q_s=\dfrac{8\times 10^3}{0.75}[/tex]

[tex]Q_s=10.667\ kJ[/tex]

Work Produced by  cycle

[tex]W=Q_s-Q_r[/tex]

[tex]W=10.667-8[/tex]

[tex]W=2.667\ kJ[/tex]

Time interval for which power is supplied

[tex]P\times t=W[/tex]

[tex]t=\dfrac{W}{P}[/tex]

[tex]t=\dfrac{2.667}{5}[/tex]

[tex]t=0.5334\ s[/tex]  

The energy taken in each cycle of this particular heat engine is 2.0 x 10^4 J, and the time interval for each cycle is four seconds.

The efficiency of a heat engine (e) is derived from work output (W) divided by the energy input (Qin). Given the mechanical power output of the engine and its efficiency, we can use this formula to determine the energy input and the time interval for each cycle.

(a) Energy taken in during each cycle: Since efficiency e = Wout/Qin, then Qin = Wout / e = 5,000W (or 5 x 103 J/s) / 0.25 = 2.0 x 104 J. Where Wout is 5000W converted to Joules per second.

(b) Time interval for each cycle: The energy balance for one cycle is given by Qin = Wout + Qexhaust, where Wout is work output and Qexhaust is exhaust energy. The time for one cycle t = Qin / W = 2.0 x 104 J / 5,000 J/s = 4 seconds. The time interval for each cycle is four seconds.

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Answer:

Angular displacement=2π/3.9 rad

Explanation:

Given data

Time t=3.9s

Required

The angular displacement during a 1.0 s time interval

Solution

In 3.9 second the child covers a full circle=2π rad

Angular displacement after 1.0 second is given as:

[tex]=\frac{2\pi }{3.9} rad[/tex]

Angular displacement=2π/3.9 rad

The child's angular displacement on the merry-go-round during a 1.0s time interval would be approximately 1.609 radians.

To find the child's angular displacement on the merry-go-round, we first need to know the rate at which the merry-go-round is turning. This is called the angular velocity and is measured in radians per seconds (rad/s). If it takes 3.9 seconds for the merry-go-round to make one full revolution, this equals 2π radians. Therefore, the angular speed of the merry-go-round is 2π/3.9 rad/s.

Now, if we want to know how much the child displaces in 1.0 second, we simply multiply the angular speed by the time interval. So the angular displacement is (2π/3.9 rad/s)*1.0s = 1.609 rad.

Therefore, the child's angular displacement during a 1.0s time interval would be approximately 1.609 radians.

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Complete Question:

Devon has several toy car bodies and motors. The motors have the same mass, but they provide different amounts of force, as shown in this table.  

The bodies have the masses shown in this table (refer attached figure).  

Which motor and body should Devon use to build the car with the greatest acceleration?

motor 1, with body 1

motor 1, with body 2

motor 2, with body 1

motor 2, with body 2

Answer:

Devon should build the car with motor 2 and body 1 for having the greatest acceleration.

Explanation:

As per Newton's second law of motion, the acceleration of any object is directly proportional to the force on the object and inversely proportional to the mass of the object.

It can be seen that motor 2 has greater force than the force provided by motor 1. Similarly, the mass of body 1 is found to be lesser compared to mass of body 2. So,

          [tex]acceleration =\frac{\text { Force }}{\text { mass }}[/tex]

It gives, the system with motor 2 and body 1 the maximum acceleration. So the car should be built with motor 2 and body 1.

The car with the greatest acceleration will be one that has a higher power-to-weight ratio, provided by a light yet powerful motor, and a lighter, aerodynamically efficient body. Both factors--lightweight and power-- are crucial for achieving high acceleration.

In deciding what motor and body to use, Devon must consider factors like the power to weight ratio, the torque of the motor, and the aerodynamics of the body. A higher power-to-weight ratio generally translates to greater acceleration. Therefore, Devon should choose a motor that is powerful yet light. When considering the body, Devon should go for a lighter body as heavy bodies slows down a car's acceleration. Besides weight, a body whose design is aerodynamically efficient will enhance acceleration because it reduces air resistance.

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Answer:

45 C.

Explanation:

Given:

Time, t = 5.0 h

= 18000 s

Current, I = 2.5 mA

= 0.0025 A

Potential difference, V = 9 V

Q = I × t

= 18000 × 0.0025

= 45 C.

Answer:

The cable would stretch 14 cm when loaded with 1000 kg ore.

Explanation:

The question is incomplete.

The complete question would be.

A mine shaft has an ore elevator hung from a single braided cable of diameter 2.5 cm. Young's modulus of elasticity of the cable is [tex]10\times 10^{10}\ N/m^2[/tex]. When the cable is fully extended, the end of the cable is 700 m below the support.

How much does the fully extended cable stretch when 1000 kg of ore is loaded into the elevator?

Given the diameter of the cable is [tex]2.5\ cm[/tex]. The length of the cable is [tex]700\ m[/tex].

And the mass of the ore is [tex]1000\ kg[/tex]. Also the Young's modulus of  elasticity of the cable is [tex]10\times 10^{10}\ N/m^2[/tex].

We will use Hook's law

[tex]\sigma=E\epsilon[/tex]

Where [tex]\sigma[/tex] is stress. E is Young's modulus of elasticity. And [tex]\epsilon[/tex] is strain.

We can rewrite .

[tex]\frac{P}{A}=E\times \frac{\delta l}{l}[/tex]

Where [tex]P[/tex] is the applied force, [tex]A[/tex] is the area of the cross-section. [tex]\delta l[/tex] is the change in length. [tex]l[/tex] is the initial length of the cable.

Also, the applied force [tex]P[/tex] is due to mass of the ore. That would be [tex]P=mg\\P=1000\times 9.81\ N[/tex]

Given diameter of the cable [tex](d)[/tex] [tex]2.5\ cm[/tex].

[tex]d=\frac{2.5}{100}=0.025\ m\\ A=\frac{\pi}{4}d^2\\ \\A=\frac{\pi}{4}(0.025)^2=4.91\times 10^{-4}\ m^2[/tex]

[tex]E=10\times 10^{10}\ N/m^2[/tex]

[tex]l=700\ m[/tex]

Plugging these values

[tex]\frac{P}{A}=E\times \frac{\delta l}{l}[/tex]

[tex]\frac{P}{A}\times \frac{l}{E}=\delta l \\\\ \delta l =\frac{1000\times 9.81\times 700}{4.91\times 10^{-4}\times 10\times 10^{10}} \\\delta l=.139\ m\\\delta l=14\ cm[/tex].

So, the cable would stretch 14 cm when loaded with 1000 kg ore.

Answer:

C

Explanation:

Both technicians are correct.

Cheers

Technician B is correct because the ECT and IAT should read similarly after a car has been sitting for a while, reflecting the same ambient temperature. TSBs can address a variety of issues, not necessarily involving a specific DTC as Technician A suggests.

Technician B is correct. Technical Service Bulletins (TSBs) typically address widespread problems or issues found in a particular model of a vehicle. While these bulletins may include information about specific Diagnostic Trouble Codes (DTCs), they do not always involve a specific stored DTC, as Technician A suggests. TSBs can cover various issues, including non-DTC-related performance, noise, and other iterative improvements that are not related to any stored codes.

Technician B is accurate in saying that the Engine Coolant Temperature (ECT) sensor and the Intake Air Temperature (IAT) sensor readings should be close to the same after the vehicle has been sitting for several hours, typically overnight. This is because both the coolant in the engine and the air in the intake manifold would have reached ambient temperature, reflecting similar temperatures if the vehicle's sensors are functioning correctly.

Answer choice B, which credits Technician B, is the correct option.

Answer:

so we have a good place to live at.

Explanation:

Environment is understood to be the set of natural and human factors that surround man in his daily life. Thus, landforms, natural resources, buildings, etc., are part of it.

Taking these definitions into account, a regular environment is one in which all its conditions and components are found with the fewest possible alterations, or with human alterations that do not negatively affect its natural conditions.

In this way, an environment that has not definitively consumed its resources or that has not significantly affected the natural status of the region is considered regular. This characteristic is important because it allows the environment to not be negatively affected, allowing a normal development of human life.

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Answer:

1. The magnetic field encircles the wire in a counterclockwise direction

Explanation:

When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.

Ever current carrying conductor produces a magnetic field around it.

Answer:

The hang time is 0.95 s

Explanation:

Spud Webb has a height of 5 feet 8 inches and is one of the shortest basketball players to play in the NBA.

He could make an amazing vertical leap of 1.1 m off the ground. For such a leap to calculate the hang time(the time spent in the air after leaving the ground and before touching down again), we use the formula:

[tex]S=ut -\frac{1}{2}gt^{2}[/tex]

Where S is the distance traveled, u is the initial velocity, t is the time taken and g is the acceleration due to gravity.

Given that:

g = 9.8 m/s²

S = [tex]y_{f} -y_{i} =0-1.1=-1.1[/tex]

u = 0

[tex]S=ut -\frac{1}{2}gt^{2}[/tex]

Substituting values:

[tex]-1.1=(0)t -\frac{1}{2}(9.8)t^{2}\\-1.1=0-4.9t^{2} \\-1.1=-4.9t^{2}[/tex]

Dividing through by -4.9 we get:

[tex]\frac{-1.1}{-4.9} =\frac{-4.9}{-4.9}t^{2}[/tex]

[tex]t^{2}=0.2245\\ t=\sqrt{0.2245}=0.474[/tex]

t = 0.474 s

The hang time = 2t = 2 × 0.474 = 0.95 s

The hang time is 0.95 s

The average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.

Explanation:

Average speed is the measure of total distance covered at different time period. Since, the formula used for calculating the average speed is the ratio of total distance covered by each car to the time taken to cover that distance.

As there is only one set of data i.e., distance and time, the average speed will be equal to the speed of the car.

So in this case, the total distance covered by franks car is 300 cm = 3 m in 1.5 s. Then, the average speed will be

[tex]Avg. S_{frank} = \frac{Total\ distance\ rolled\ by\ his\ car}{Time\ taken}[/tex]

[tex]Avg.S_{frank} =\frac{300 \times 10^{-2} }{1.5}=2\ m/s[/tex]

Similarly, the average speed of Noahs car which rolled a distance of 360 cm = 3.6 m in time 2 s, will be

[tex]Avg.S_{Noah} =\frac{360 \times 10^{-2} }{2}=1.8\ m/s[/tex]

Thus, the average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.

The average speed for Frank's car and Noah's car is 200 cm/s and 180 cm/s respectively calculated using the formula speed = distance / time.

Average speed is a measure of how quickly an object moves over a certain distance in a specific amount of time. It's calculated using the formula:

Average Speed (v) = Total Distance Traveled (d) / Total Time Taken (t)

To calculate the average speed of any moving object, use the formula: speed = distance / time. For Frank's car which rolled 300 cm in 1.5 seconds, the speed would be 300 cm / 1.5 s = 200 cm/s. Similarly, for Noah's car which rolled 360 cm in 2 seconds, the speed would be 360 cm / 2 s = 180 cm/s.

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Final answer:

Total internal reflection is possible for light incident from air on water.

Explanation:

Total internal reflection occurs when light traveling from a medium with a higher refractive index to a medium with a lower refractive index is incident at an angle greater than the critical angle.

In the case of light incident from air on water, the critical angle is 48.6°. If the angle of incidence is greater than this critical angle, the light will undergo total internal reflection and not refract out of the water. Therefore, total internal reflection is possible for light incident from air on water.

Answer:

A=0

Explanation:

I got 100% on this assignment

This question involves the basic concept of displacement.

The frog's displacement is "0 cm".

In order to calculate the displacement of the frog, we must consider both the magnitude and direction of the movements made by the frog. Here, we will take the forward direction as positive and the backward direction as negative. Hence, the displacement of the frog will be as follows:

Displacement = 18 cm - 6 cm - 12 cm

Displacement = 0 cm

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The attached picture shows the difference between the displacement and the distance.

Explanation:

Answer:Hydrogen lines will be weak if the star is too hot or too cold.

Explanation:

At higher temperatures, the hydrogen atom ionizes due to the atomic collisions. M spectral class stars are mainly the main sequence and red stars. They are in a temperature range of 3000 K which means that these stars have maximum ionized calcium lines. In this star, hydrogen atoms have electrons in the lower energy state and it is difficult to absorb photons. These stars do not have enough temperature for absorption and undergo fusion.

Final answer:

The force of gravity and air resistance do work on a simple pendulum, with gravity doing positive and negative work as the pendulum moves, and air resistance dissipating energy from the system. Tension in the string does not do work since it is always perpendicular to the pendulum's motion.

Explanation:

In the context of a simple pendulum, the forces capable of doing work are the force of gravity and air resistance. The force of gravity does work on the pendulum as it swings back and forth because it has a component along the direction of the pendulum's movement. Specifically, as the pendulum swings, gravity pulls it downward, causing it to accelerate towards the lowest point of its path. Upon reaching this lowest point, the pendulum has maximum kinetic energy because the potential energy due to its elevated position has been converted into kinetic energy. The pendulum then slows down as it climbs against gravity until it stops momentarily at the highest point of its swing, and gravity starts doing negative work, converting kinetic energy back into potential energy. The tension in the string, although a force present in the pendulum system, does no work because it acts perpendicular to the direction of the pendulum's motion at all times.

Air resistance does work, although typically very small, by opposing the motion of the pendulum and thus removing energy from the system, mostly in the form of thermal energy due to the friction between the air and the pendulum bob. In simple pendulum motion, air resistance is usually considered negligible, but it does contribute to the eventual stopping of the pendulum due to energy dissipation.

The force of gravity does work in both directions, the tension in the string does no work, and air resistance does work by dissipating the pendulum's energy.

Force of Gravity (Weight): The force of gravity does work on the pendulum bob. Gravity acts downward and provides the restoring force that accelerates the pendulum back toward its equilibrium position. As the pendulum swings, gravity causes the bob to move, thereby doing positive work as the bob descends and negative work (taking energy away from the pendulum's motion) as the bob ascends.

Tension in the String: The tension in the string does no work on the pendulum bob. This is because the tension force always acts perpendicular to the direction of the bob's instantaneous motion. Work is defined as the force component in the direction of displacement times the displacement itself. Since there is no displacement in the direction of the tension force, it does no work.

Air Resistance: Air resistance does work on the pendulum bob, but in the opposite sense to gravity. Air resistance acts against the direction of motion of the bob, causing it to lose energy over time. This dissipation of energy due to air resistance results in damping, gradually reducing the amplitude of the pendulum's oscillation.

Answer:

lead

Explanation:

Lead was once widely used in the United States as a gasoline additive.

Addition of lead is in the form of tetra ethyl lead(II).

It helps to improve the octane rating of gasoline and to produce more useful energy via each combustion step.

Answer:

The displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.

Explanation:

Given:

Distance moved in the East direction (d) = 45 km

Distance moved in the North direction (D) = 110 km

Displacement is defined as the difference of final position and initial position.

Let us draw a diagram representing the above situation.

Point A is the starting point and point C is the final position of RV.

So, the displacement of RV in two days is given as:

Displacement = Final position - Initial position = AC

Now, triangle ABC is a right angled triangle with AB = 45 km, BC = 110 km, and AC being the hypotenuse.

Using Pythagoras theorem, we have:

[tex]AC^2=AB^2+BC^2\\\\AC=\sqrt{AB^2+BC^2}[/tex]

Plug in the given values and solve for AC. This gives,

[tex]AC=\sqrt{(45\ km)^2+(110\ km)^2}\\\\AC=\sqrt{2025+12100}\ km\\\\AC=\sqrt{14125}\ km\\\\AC=118.85\ km[/tex]

Now, the direction of displacement with the east direction is given as:

[tex]\theta=\tan^{-1}(\frac{BC}{AB})\\\\\theta =\tan^{-1}(\frac{110}{45})=67.75^\circ[/tex]

Therefore, the displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.

The displacement of the rv over the two days is 118.85 km.

The given parameters;

The displacement of the rv over the two days is calculated by applying Pythagoras theorem as follows;

[tex]c^2 = a^2 + b^2\\\\c = \sqrt{a^2 + b^2} \\\\c = \sqrt{(45)^2 + (110)^2} \\\\c = 118.85 \ km[/tex]

Thus, the displacement of the rv over the two days is 118.85 km.

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Answer:

[tex]E_a=1124.83 J/mol[/tex]

Explanation:

Given that second order equation

K₁ = 0.0796 L/mol-s , T₁= 737⁰C

T₁ = 737 + 273 K = 1010 K

K₂ = 0.0815 L/mol-s , T₂=947°C

T₂=947+273 K= 1220 K

The activation energy given as follows

[tex]\ln\dfrac{K_2}{K_1}=\dfrac{E_a}{R}\left ( \dfrac{1}{T_1}-\dfrac{1}{T_2} \right )[/tex]

Now by putting the values we can get

[tex]\ln\dfrac{0.0815}{0.0796}=\dfrac{E_a}{8.314}\left ( \dfrac{1}{1010}-\dfrac{1}{1220} \right )[/tex]

[tex]0.023=0.00017\times \dfrac{E_a}{8.314}[/tex]

[tex]E_a=0.023\times \dfrac{8.314}{0.00017}[/tex]

[tex]E_a=1124.83 J/mol[/tex]

Therefore the activation energy will be 1124.83 J/mol

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

[tex]F=6.00*10^{-6}N[/tex]

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

[tex]F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}[/tex]

Substitute the given values

[tex]F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N[/tex]

Both the objects travel equal distance before stopping.

Explanation:

Given-

Mass of object 1, m₁ = 4kg

Speed of object 1, v₁ = 2m/s

Mass of object 2, m₂ = 1kg

Speed of object 2, v₂ = 4m/s

Force₁ = Force₂ = F

Distance, s = ?

We know,

[tex]v^2 - u^2 = 2as\\\\[/tex]

Where, v is the final velocity

            u is the initial velocity

            a is the acceleration

            s is the distance

When the brake is applied, the object comes to rest and the final velocity, v becomes 0. So,

[tex]s = \frac{u^2}{2a}[/tex]

We know,

[tex]a = \frac{F}{m}[/tex]

The stopping distance becomes,

[tex]s = \frac{u^2m}{2F}[/tex]

For object 1:

[tex]s = \frac{(2)^2 X 4 }{F}[/tex]

[tex]s = \frac{16}{F}[/tex]

For object 2:

[tex]s = \frac{(4)^2 X 1}{F}\\ \\s = \frac{16}{F}[/tex]

For both the objects the distance travelled is same.

Therefore, both the objects travel equal distance before stopping.

Answer:

Center of mass

Explanation:

The Doppler technique is a good method for discovering exoplanets. It uses the Doppler effect to analyze the motion and properties of the star and planet. Both the planet and the star are orbiting a common center of mass. This means that the star and the planet gravitationally attract one another, causing them to orbit around a point of mass central to both bodies.

If we use the Doppler method to measure the period with which a star alternately moves toward and away from us due to an orbiting planet, then we also know the planet's orbital period

When utilizing the Doppler method to measure the periodic shifts in a star's spectral lines caused by an orbiting planet, not only do we discern the star's radial velocity variations, but crucially, we also ascertain the orbital period of the planet. This period signifies how long it takes for the planet to complete one orbit around its host star.

The Doppler effect causes a star's light to shift towards the blue end of the spectrum as it moves closer to Earth and towards the red end as it moves away. By analyzing these shifts, astronomers can deduce the star's motion induced by the gravitational pull of its companion planet. The periodicity of these shifts corresponds directly to the planet's orbital period. This information is fundamental in understanding the planet's characteristics, including its distance from the star and its mass, contributing significantly to our knowledge of exoplanetary systems in the vast cosmos.

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Answer:

[tex]t_t=4.131\ s[/tex]

Explanation:

Given:

height above the horizontal form where the ball is hit, [tex]y=1\ m[/tex]

angle of projectile above the horizontal, [tex]\theta=30^{\circ}[/tex]

initial speed of the projectile, [tex]u=40\ m.s^{-1}[/tex]

Firstly we find the vertical component of the initial velocity:

[tex]u_y=u.\sin\theta[/tex]

[tex]u_y=40\times \sin30^{\circ}[/tex]

[tex]u_y=20\ m.s^{-1}[/tex]

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

Using eq. of motion:

[tex]v_y^2=u_y^2-2g.h[/tex] (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

[tex]0^2=20^2-2\times 9.8\times h[/tex]

[tex]h=20.4082\ m[/tex] (from the height where it is thrown)

Now we find the time taken to ascend to this height:

[tex]v_y=u_y-g.t[/tex]

[tex]0=20-9.8t[/tex]

[tex]t=2.041\ s[/tex]

Time taken to descent the total height:

[tex]h'=u_y'.t'+\frac{1}{2} g.t'^2[/tex]

[tex]21.4082=0+4.9t'^2[/tex]

[tex]t'=2.09\ s[/tex]

Now the total time taken by the ball to hit the ground:

[tex]t_t=t'+t[/tex]

[tex]t_t=2.09+2.041[/tex]

[tex]t_t=4.131\ s[/tex]

Answer:

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given that,

The radius of earth r = 6.3781×10^6 m

The angular speed of earth w = 7.27*10^-5 rad/s

The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

The mass of earth m = 5.972 × 10^24 kg

The inertia of sphere I = 2/5 * m* r^2

Solution:

angular acceleration of the earth from rest to w is given by α:

                               α = w / t

                               α = (7.27*10^-5) / (1.57784760 * 10^8)

                               α = 4.60754*10^-13 rad/s^2

The required torque τ is given by:

                               τ = I*α

                               τ = 2/5 * m* r^2 * α

 τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

 τ = 4.47746 * 10^25 N-m

Power required P to turn the earth to the speed w is:

                          P = τ*w

                          P = (4.47746 * 10^25)*(7.27*10^-5)

                          P = 3.25511*10^21 W

Energy E required is :

                          E = P / t

                          E = (3.25511*10^21) / (1.57784760 * 10^8)

                          E = 2.06301 * 10^13 J

Answer:

Orbital period of satellite is 5.83 x 10³ s

Explanation:

The orbital period of satellite revolving around Earth is given by the equation :

[tex]T=\sqrt{\frac{4\pi ^{2} (R+h)^{3} }{GM} }[/tex]      .....(1)

Here R is radius of Earth, h is height of satellite from the Earth's surface, M is mass of Earth and G is gravitational constant.

In this problem,

Height of satellite, h = 500 km = 500 x 10³ m

Substitute 6378.1 x 10³ m for R, 500 x 10³ m for h, 5.972 x 10²⁴ kg for M and 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻² for G in equation (1).

 [tex]T=\sqrt{\frac{4\pi ^{2} [(6378.1+500)\times10^{3} ]^{3} }{6.67\times10^{-11} \times5.972\times10^{24} } }[/tex]

T = 5.83 x 10³ s