A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.

Answers

Answer 1

Answer:

The density of the mixture is 0.55kg/m^3

Explanation:

P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole

PV = nRT

V = nRT/P = 1×8.314×273/100 = 22.70m^3

Mass of N2 = 0.4×28 = 11.2kg

Mass of H2 = 0.6×2 = 1.2kg

Mass of mixture = 11.2 + 1.2 = 12.4kg

Density of mixture = mass/volume = 12.4/22.7 = 0.55kg/m^3


Related Questions

A person is riding on a Ferris wheel. When the wheel makes one complete turn, the net work done on the person by the gravitational force ____________.

Answers

Answer:

0

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

d = Vertical height from the ground

F = Force = Weight = mg

Net work done would be

[tex]W_n=W_{up}+W_{down}\\\Rightarrow W_n=Fdcos180+Fdcos0\\\Rightarrow W_n=-mgd+mgd\\\Rightarrow W_n=0[/tex]

Hence, the work done on the person by the gravitational force is 0

The maximum force a pilot can stand is about seven times his weight. What is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250 m/s?

Answers

Answer:

Radius of curvature of the path is 1063 meters

Explanation:

It is given that,

Force acting on the pilot is about seven times of his weight. Speed with which pilot moves, v = 250 m/s.

As per Newton's second law of motion, the net force acting on the pilot at the bottom is given by :

[tex]N-mg=\dfrac{mv^2}{r}[/tex]

Where

N is the normal force

r is the radius of curvature

According to given condition,

[tex]7mg-mg=\dfrac{mv^2}{r}[/tex]    

[tex]6mg=\dfrac{mv^2}{r}[/tex]    

[tex]r=\dfrac{mv^2}{6mg}[/tex]  

[tex]r=\dfrac{mv^2}{6mg}[/tex]      

[tex]r=\dfrac{v^2}{6g}[/tex]      

[tex]r=\dfrac{(250)^2}{6\times 9.8}[/tex]  

r = 1062.92 meters

or

r = 1063 meters

So, the radius of curvature of the path is 1063 meters. Hence, this is the required solution.

A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A transit train left from station A 1 hour later than the passenger train, but it arrived at the station B at the same time with the passenger train. What was the average speed of the transit train?

Answers

Average speed of transit train is 60 mph

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train  = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train  =  4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.
a) Through how many radians must the spool be turned to wrap one even layer of wire?
b) What is the length of this wound wire?

Answers

Answer:

a)   # lap = 301.59 rad , b)   L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

          # turns = 1 turn (24 cm / 0.5 cm)

         # laps = 48 laps

Let's reduce to radians

        # laps = 48 laps (2 round / 1 round)

       # lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

          L₀ = 2π R

          L = # turns L₀

          L = # turns 2π R

          L = 48 2π 30

          L = 9047.79 cm

          L = 90.48 m

The spool needs to be turned through 2π radians to wrap one even layer of wire. The length of the wound wire can be calculated using the circumference formula.

Part a: To wrap one even layer of wire around the spool, the spool must be turned through an angle of 2π radians because one complete revolution is equal to 2π radians.

Part b: To calculate the length of the wire wound on the spool, we find the circumference of the spool using the formula: circumference = 2πr, where r is the radius of the spool (30 cm). The length of the wound wire is the product of this circumference and the height of the spool (24 cm).

Why do astronauts (and cans of soft drink) float around in the shuttle instead of falling?

Answers

Answer: Astronauts only float around in the shuttle when they are outside the gravitational pull of the earth

Explanation: when astronauts takes off from the earth, they get to a point (space) where the earth's gravity can no longer pull them. At this state, they experience weightlessness because there is no gravity. Since there is no gravity to pull them down, hence they start floating.

Final answer:

Astronauts and objects float in the shuttle due to the microgravity state from continual free-fall around the Earth, creating the feeling of weightlessness.

Explanation:

Astronauts and objects like cans of soda float in the shuttle because of the lack of gravity in space, a state known as microgravity. When the shuttle is orbiting the earth, it's actually falling towards the earth but also moving forward. This forward motion allows the shuttle and everything in it to keep missing the Earth, so they keep falling towards it but never hitting it. This continual state of free-fall creates the feeling of weightlessness and is why astronauts and objects inside the shuttle appear to float.

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A Styrofoam cooler has outside dimensions of B = 84.0 cm, W = 47.0 cm, and H = 59.0 cm. The thickness of each wall of the cooler, t, is 8.0 cm. Styrofoam has a density, rho, of 1.0 kg/m3. (a) What is the volume of the Styrofoam used in cubic inches? (b) What is the mass in lbm? (c) How many gallons of liquid could be stored in the cooler?

(a) The volume of the Styrofoam used is 14214.37 in3.
(b) The mass is 9.6 lbm.
(c) The number of gallons of liquid that could be stored in the cooler is 1058.54 gal.

Answers

Answer:

a) [tex]V_s=4989.7895\ in^3[/tex]

b) [tex]mass=0.1803\ lb[/tex]

c) [tex]V_i=39.93\ gallons[/tex]

Explanation:

external breadth of cooler, [tex]B=84\ cm[/tex]external width of cooler, [tex]W=47\ cm[/tex]external height of cooler, [tex]H=59\ cm[/tex]∵ thickness of each wall is, [tex]t=8\ cm[/tex]

Therefore,

internal breadth of cooler, [tex]B_i=76\ cm[/tex]internal width of cooler, [tex]W_i=39\ cm[/tex]internal height of cooler,  [tex]H_i=51\ cm[/tex]

a)

External volume of the structure:

[tex]V=B.H.W[/tex]

[tex]V=84\times47\times59\div 2.54^3[/tex]

[tex]V=14214.3828\ in^3[/tex]

Internal volume of the structure:

[tex]V_i=B_i.H_i.W_i[/tex]

[tex]V_i=76\times 39\times 51\div 2.54^3[/tex]

[tex]V_i=9224.5932\ in^3[/tex]

∴Volume of Styrofoam used:

[tex]V_s=V-V_i[/tex]

[tex]V_s=14214.3828-9224.5932[/tex]

[tex]V_s=4989.7895\ in^3[/tex]

b)

given that density of Styrofoam, [tex]\rho=1\ kg.m^{-3}=3.613\times 10^{-5}\ lb.in^{-3}[/tex]

we know,

[tex]\rm mass= density \times volume[/tex]

[tex]mass=4989.7895\times 3.613\times 10^{-5}[/tex]

[tex]mass=0.1803\ lb[/tex]

c)

Volume of liquid it can hold [tex]=V_i[/tex]

[tex]V_i=39.93\ gallons[/tex]

Final answer:

The volume of the Styrofoam used is 14214.37 in³. The mass is 9.6 lbm. The number of gallons of liquid that could be stored in the cooler is 1058.54 gal.

Explanation:

To find the volume of the Styrofoam used in cubic inches, we need to convert the outside dimensions from centimeters to inches, subtract the volume of the cooler, and divide by the thickness of each wall. The volume of the Styrofoam used is 14214.37 in³.

To find the mass in lbm, we need to convert the density from kg/m³ to lbm/in³ and multiply by the volume of the Styrofoam used. The mass is 9.6 lbm.

To find the number of gallons of liquid that could be stored in the cooler, we need to convert the volume of the cooler from cm³ to gallons. The number of gallons of liquid that could be stored in the cooler is 1058.54 gal.

to obtain a multimeter reading of 1 v between b and c in the circuit the value of r2 would have to be.

Answers

Complete question:

In the circuit shown in the figure below (See image attached), suppose that the value of R1 is [tex] 500\,k\Omega [/tex]. To obtain a multimeter reading of 1 V between points B and C in the circuit, the value of R2 would have to be.

Answer:

[tex]R2=0.1\Omega [/tex]

Explanation:

First, we are going to the find current trough the circuit, because the resistors are on series the current is the same on each resistor so I=I1=I2. The Ohm's law for the circuit is:

[tex] V=R_{T}*I [/tex] (1) , with V the voltage of the battery (6V), I the current trough the circuit and [tex]R_{T} [/tex] the total resistance of the circuit, but for resistors on series the total resistance is the sum of the individual resistance so [tex] R_{T} = R1+R2[/tex] (2).

Using (2) on (1) and solving for I:

[tex]I=\frac{V}{R1+R2} [/tex] (3)

Ohm's law is true for the individual resistors too so we're going to apply that on R2:

[tex]V2=R2*I2 [/tex], but remember I2=I

[tex]V2=R2*I [/tex] (4), using (3) on (4)

[tex]V2=R2* \frac{V}{R1+R2} [/tex], solving for R2:

[tex] R1*V2+R2*V2=R2V[/tex]

[tex] R1*V2=R2(V-V2)[/tex]

[tex] R2=\frac{R1V2}{V-V2}=\frac{500\times10^{3}\Omega*1V}{6V-1V}[/tex], V2= 1V because we want that reading on the multimeter.  

[tex]R2=100\,k\Omega [/tex]

Answer:

10k

Explanation:

An insulating, solid sphere has a uniform, positive charge density of [tex]\rho[/tex] =4.40*10⁻⁷C/m³. The sphere has a radius R of 0.370m.
What is the potential [tex]V_r[/tex] at a point located at r = 0.160 m from the center of the sphere?

Answers

Final answer:

The electric potential at a point within a uniformly charged solid sphere can be calculated using the given formula. By substituting the given values of charge density, radius of the sphere, and distance from the center of the sphere, the electric potential can be computed.

Explanation:

The potential Vr within a uniformly charged solid sphere of charge density ρ, radius R, at a distance r from the center is given by the formula:

Vr = ρr²/6ε₀ + (ρR²/2ε₀) [ 1 - 3r²/R² ] for r < R

Where ρ is the charge density (4.40*10⁻⁷ C/m³), r is the distance from the center of the sphere (0.160 m), R is the radius of the sphere (0.370 m) and ε₀ is the permittivity of free space (8.854 * 10⁻¹² C²/N.m²).

Substituting the given values into the formula, we can calculate the electric potential at a point located at r = 0.160 m from the center of the sphere.

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The potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]

The potential  at a point located at a distance \(r\) from the center of a uniformly charged solid sphere can be calculated using the formula:

[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]

 where \(k\) is the Coulomb's constant ([tex]\(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)), \(Q_{enc}[/tex]\) is the charge enclosed within the radius \(r\), and \(r\) is the distance from the center of the sphere to the point where the potential is being calculated.

 For a uniformly charged sphere, the charge enclosed within a radius \(r\) is given by:

[tex]\[ Q_{enc} = \frac{4}{3}\pi r^3 \rho \][/tex]

where [tex]\(\rho\)[/tex] is the charge density of the sphere.

 Given that [tex]\(\rho = 4.40 \times 10^{-7} \, \text{C/m}^3\)[/tex] and [tex]\(r = 0.160 \, \text{m}\),[/tex] we can calculate \(Q_{enc}\) as follows:

[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.160 \, \text{m})^3 (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]

[tex]\[ Q_{enc} = \frac{4}{3}\pi (0.004096 \, \text{m}^3) (4.40 \times 10^{-7} \, \text{C/m}^3) \][/tex]

[tex]\[ Q_{enc} = \frac{4}{3}\pi (1.8192 \times 10^{-9} \, \text{C}) \][/tex]

[tex]\[ Q_{enc} = 2.42528 \times 10^{-9} \, \text{C} \][/tex]

 Now, we can calculate the potential [tex]\(V_r\):[/tex]

[tex]\[ V_r = \frac{kQ_{enc}}{r} \][/tex]

[tex]\[ V_r = \frac{(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) (2.42528 \times 10^{-9} \, \text{C})}{0.160 \, \text{m}} \][/tex]

[tex]\[ V_r = \frac{2.17836 \times 10^{-8} \, \text{Nm}^2/\text{C}}{0.160 \, \text{m}} \][/tex]

[tex]\[ V_r = 1.36148 \times 10^{-7} \, \text{V} \][/tex]

 Therefore, the potential at a point located at [tex]\(r = 0.160 \, \text{m}\)[/tex] from the center of the sphere is approximately [tex]\(1.36 \times 10^{-7} \, \text{V}\).[/tex]

When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials A) conduct electricity, while others do not. B) can accept protons more easily than others. C) can give up electrons more readily than others. correct D) repel negative objects, and others attract them.

Answers

When a balloon is rubbed with human hair, the balloon acquires an excess static charge. This implies that some materials can give up electrons more readily than others.

Answer: Option C

Explanation:

We know that charges can neither be created nor be destroyed by law of conservation of charges. So when we rub two objects, it is natural to have a transfer of charges. But the charges which get transferred may be negligible in most of the cases leading to no significant observations.

But for some materials, like when we rubbed a balloon with human hair, we observed clouding of excess static charge on the balloon surface. This indicates that hair can easily give up electrons to balloon leading to clouding of excess static charge on it.

What happens when a main-sequence star exhausts its core hydrogen fuel supply?

Answers

Explanation:

when a main-sequence star exhausts its core hydrogen fuel supply the core starts to shrink ( lack of fusion reactions) and the rest of the star starts to expands. The fusion reaction leaves the main sequence and begin to fuse helium in a shell outside the core. This mass stars become red supergiant  and then evolve to become blue super giant.

Answer:  Their temperature decreases dramatically, but their luminosity increases only slightly.

Explanation:  This is exact from Plato

Which of the following is characteristic of proficient catching?

A. pointing the fingers upward to catch a high ball
B. pointing the fingers upward to catch a low ball
C. letting the shoulders and elbows move to give with contact of the ball
D. A and C E. B and C

Answers

Answer:

The correct answer is option D i.e. A and C

Explanation:

The correct answer is option D i.e. A and C

for proficient catching player must

- learn to absorbed the ball force

- moves the hang according to ball direction to hold the ball

- to catch ball at high height move the finger at higher position

- to catch ball at low height move the finger at lower position

A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by a current flowing southward at 2.0 meter a second if he ended up exactly across the stream from where he began at what angel to the shore must he swim upstream

Answers

Answer:

Angle: [tex]48.19^o[/tex]

Explanation:

Two-Dimension Motion

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

[tex]v_b\ cos\alpha[/tex]

where [tex]v_b[/tex] is the speed of the boy in still water and [tex]\alpha[/tex] is the angle respect to the shoreline. If the river flows at speed [tex]v_s[/tex], we now set

[tex]v_b\ cos\alpha=v_s[/tex]

[tex]\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}[/tex]

[tex]\alpha=48.19^o[/tex]

Through which one of the following mediums is the velocity of a sound wave the greatest?

A. Steel
B. Water
C. Air
D. Lead

Answers

Answer: C. Steel

Explanation: When a sound wave travels through a solid body consisting

of an elastic material, the velocity of the wave is relatively

high. For instance, the velocity of a sound wave traveling

through steel (which is almost perfectly elastic) is about

5,060 meters per second. On the other hand, the velocity

of a sound wave traveling through an inelastic solid is

relatively low. So, for example, the velocity of a sound wave

traveling through lead (which is inelastic) is approximately

1,402 meters per second.

The closure temperature represents the point when ________. A. a magma cools to the point where minerals begin to crystallize B. radioactive isotopes begin to decay C. isotopes are no longer free to move out of a crystal lattice D. the last remaining magma crystallizes

Answers

The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice.

Answer: Option C

Explanation:

The closure temperature can also be termed as blocking temperature. It is mostly used in radiometric dating. As the temperature decreases, below a certain point the isotopes may get freeze in their lattice positions. And there may be slowing of diffusion.

At the closure temperature, that rate of diffusion will be zero as the isotopes will be no longer free to move out of crystal lattice. So, this is termed as closure or blocking temperature. As the isotopes loose their ability to move, their concentration will remain fixed in their position leading to measurement of radiation dating.

Final answer:

Closure temperature is the point where isotopes are trapped within a crystal lattice, marking the mineral as a closed system, which is crucial for isotopic dating.

Explanation:

The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice. This is a point in the cooling process of a rock or mineral, after crystallization, at which its isotopic constituents are no longer able to escape from its crystal lattice. Essentially, it is the temperature below which a mineral becomes a closed system for certain isotopes. Until reaching this temperature, isotopic parent and daughter products can freely exchange with the environment. As the temperature drops below the closure temperature, the system locks in place, preserving the isotopic date that will be analyzed to determine the history of the rock.

A flat, circular, metal loop of radius r = 1 m is at rest in a uniform magnetic field of magnitude B. The plane of the loop is parallel to the page and the magnetic field is directed perpendicular to and out of the page, as indicated by the blue dots. If the magnitude of the magnetic field increases from 2 T to 6 T in 2 s, what is the magnitude of the induced emf within the circular loop?

Answers

Answer:

Induced EMF,[tex]\epsilon=6.28\ volts[/tex]

Explanation:

Given that,

Radius of the circular loop, r = 1 m

Time, t = 2 s

Initial magnetic field, [tex]B_i=2\ T[/tex]

Final magnetic field, [tex]B_f=6\ T[/tex]

The expression for the induced emf within the circular loop is given by :

[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]

[tex]\phi[/tex] = magnetic flux

[tex]\epsilon=\dfrac{d(BA\ cos\theta)}{dt}[/tex]

Here, [tex]\theta=90\ degrees[/tex]

[tex]\epsilon=A\dfrac{d(B)}{dt}[/tex]

[tex]\epsilon=A\dfrac{B_f-B_i}{t}[/tex]

[tex]\epsilon=\pi (1)^2\times \dfrac{6-2}{2}[/tex]

[tex]\epsilon=6.28\ volts[/tex]

So, the induced emf in the loop is 6.28 volts. Hence, this is the required solution.  

Technician A says that the waveform produced by an ac generator after rectification is called a sine wave. Technician B Says that the waveform produced by the ac generator after the output moves through the diodes is a straight line because it is a constant dc voltage. who is correct?

A.Tech A
B.Tech b
C.A and B
D. neither a nor b

Answers

Neither Technician A nor B is correct.

Answer: Option D

Explanation:

In AC generator, diodes are usually used to rectify the alternator current. This is the transformation into direct current from alternating current. Sinusoidal voltage is generated by generators when it rotates in both directions. The output voltage [tex]V_{i n}[/tex] is converted to DC with diode help.

This process is called rectification. The output voltage [tex]V_{i n}[/tex]  is treated as the input signal of the diode circuit. So, after rectification, the sine wave is not produced by an AC generator. And, the sine wave resulted by AC generator when the output passes via diodes and not a straight line.

Final answer:

The waveform produced by an AC generator after rectification is a pulsating DC waveform, not a sine wave. Technician A is correct in this case.

Explanation:

Technician A is correct. The waveform produced by an AC generator after rectification is not a sine wave, but a pulsating DC waveform. After the output of the generator moves through the diodes, the negative half-cycles are clipped, resulting in a waveform that resembles a series of straight lines instead of a smooth sine wave. This pulsating DC waveform still contains alternating current components, but it is not a constant DC voltage.

Which of the following explains why international travelers need to use special adapters when plugging a device designed to operate in the U.S. into a wall outlet in Europe?
The increased voltage causes the device's resistance to increase, and the device won't work.
The increased voltage causes the device's resistance to decrease, and the device will over-heat.
The increased voltage causes an increase in power usage, and the device will over-heat.
The increased voltage causes a decrease in power usage, and the device won't work.

Answers

Answer:

The answer is: The increased voltage causes an increase in power usage, and the device will over-heat.

Explanation:

First, we must consider the variables of the electrical system that will allow us to respond. In this case, power, current and voltage, which are related by

[tex]P=VI[/tex]

Where P=Power, V=Voltage, I=Current.

In the equation it can be observed that power is directly proportional to the system voltage. Thus, if the voltage increases as in this case, the power will also increase, which overheats the device and can cause damage to it.

If a 4.5-meter wave traveled at a speed of 15 m/s, what would its frequency be?

Answers

Answer:

The frequency of the wave = 10 Hz.

Explanation:

Wave: A wave is a disturbance that travels through a medium a transfer energy from one point to another in the medium without causing and permanent displacement of the medium itself.

V = λf .................. Equation 1

making f the subject of the equation,

f = V/λ.................... Equation 2

Where V = velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

Given: V = 15 m/s², λ = 4.5 m.

Substituting these values into equation 2,

f = 15/1.5

f = 10 Hz.

Therefore the frequency of the wave = 10 Hz.

Answer:

The frequency of the wave = 10 Hz.

Explanation:

the guy below gave a good explanation

A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is spinning at 200 rpm and has a mass of 28 kg. The coefficient of kinetic friction between the knife and the stone is 0.20. If the stone slows steadily to 180 rpm in 10 s of grinding, what is the force with which the man presses the knife against the stone?

Answers

Answer:

F = -2.205N

Explanation:

First, we have to find the angular aceleration due to the knife following the next equation:

W = Wo + at

where W is the final angular velocity and Wo is the initial angular velocity, a the angular aceleration and t the time.

Now, we will change the angular velocity to rad/s as:

Wo = 200 rpm = 20.94 rad/s

W = 180 rpm = 18.84 rad/s

Replacing in the previus equation, we get:

18.84rad/s = 20.94rad/s + a(10s)

solving for a:

a = -0.21rad/s^2

Now, we have to find the moment of inertia of the grindstone using:

I = [tex]\frac{1}{2}MR^2[/tex]

Where M is the mass of the stone and R the radius of the stone. Replacing values:

I = [tex]\frac{1}{2}(28kg)(0.15m)^2[/tex]

I = 0.315 kg*m^2

Adittionally:

T = Ia

where T is the torque, I the moment of inertia and a the angular aceleration.

so:

[tex]U_kFd = Ia[/tex]

where [tex]U_k[/tex] is the coefficient of the kinetic friction, F is the force with which the man presses the knife and d the lever arm. So, replacing values, we get:

[tex](0.2)F(0.15m) = (0.315)(-0.21rad/s^2)[/tex]

solving for F:

F = -2.205N

it is negative because the stone is stopping due of this force.

If Mars were the same size as Mercury (instead of its actual size), which surface features would it have?

Answers

Answer:

A surface that is full of craters, very similar to that one seen in mercury.

Explanation:

Mercury is the smaller planet in the solar system whit a very thin atmosphere. If Mars were the same size of mercury it will certainly have a surface full of craters.

When meteoroid (fragments of rock) reach the atmosphere of a planet, they get incinerate as a consequence of the friction between the object and the atmosphere. However, if the meteoroid has the necessary size it can reach the ground (at this point is known as meteorite).

In the case propose, it is most likely that bigger meteorite reach the surface of the planet, which leads to the result of a surface full of craters since the size of mars is not enough to maintain a very dense atmosphere due to the weak gravitational field.

A 84.5 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astronaut turns away to look at Earth and several seconds later is 38.9 m behind the ship, at rest relative to the spaceship. The only way to return to the ship without a thruster is to throw a wrench directly away from the ship. The wrench has a mass of 0.613 kg, and the astronaut throws the wrench with a speed of 24.9 m/s. How long does it take the astronaut to reach the ship? Answer in units of s.

Answers

Answer:

215.35736 seconds

Explanation:

[tex]m_1[/tex] = Mass of astronaut = 84.5 kg

[tex]m_2[/tex] = Mass of wrench = 0.613 kg

[tex]v_1[/tex] = Velocity of astronaut

[tex]v_2[/tex] = Velocity of wrench = 24.9 m/s

In this system the linear momentum is conserved

[tex]m_1v_1=m_2v_2\\\Rightarrow v_1=\dfrac{m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{0.613\times 24.9}{84.5}\\\Rightarrow v_1=0.18063\ m/s[/tex]

Time is given by

[tex]Time=\dfrac{Distance}{Speed}[/tex]

[tex]Time=\dfrac{38.9}{0.18063}=215.35736\ s[/tex]

The time it will take the astronaut to get back to the ship is 215.35736 seconds

What are the potentially negative impacts of movement corridors?

Answers

Answer:

These corridors can decrease the inbreeding in declining populations and enhance the spread of different diseases.

Explanation:

Movement corridors generally will enhance a process such as dispersal which is the spread or distribution of things over a considerably large area. These corridors can decrease the inbreeding in declining populations. They are very vital to species that usually migrate seasonally. On the other day, the movement corridor can be highly harmful in nature due to its ability of enhancing the spread of different diseases. These corridors facilitate the movement of species and allow the spread of harmful diseases.

Help Please ! A wave of wavelength 52 cm travels with frequency 400 Hz. What is the speed of the wave? Show all work.

Answers

The speed of the wave is 208 m/s

Explanation:

The wave equation states that ; speed of wave= wavelength *frequency

Mathematically, v=λ*f

Given;

λ=52 cm= 52/100 =0.52 m

f= 400 Hz

v=?

v=0.52*400 =208 m/s

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Keyword : wave, wavelength, frequency, speed

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As a result of cancer, Mike had to have part of his cecum removed. Mike is probably at less risk for nutrient absorption problems than he would be if part of his _____ had been removed.

Answers

Answer: Small intestine

Explanation: This is because the small intestine is the place where absorption of minerals and nutrients from food takes place.

The small intestine is also known as small bowels, it is located between the large intestine and the stomach where the pancreatic duct supplies it with pancreatic juice and bile that helps digestion.

Large, heavy vehicles have slow acceleration, long__________distances and respond poorly through sharp turns.

Answers

Answer:braking

Explanation:

Large vehicles as compared to small vehicles require long braking distance, otherwise, it could topple the heavy vehicles. Heavy vehicles provide high torque thus it is used to carry heavy loads.

They run at relatively low speed as compared to the light vehicles as they are slow to accelerate and thus require long braking distance as Momentum associated with them is very high.

If sudden brakes are applied it may cause the vehicle to skid and flip over it due to the presence of large momentum.

Suppose we had two speakers, one placed some distance directly behind the other andboth pointing along the same direction, facing us. Now, we connect the speakers tothe same source and turn them on producing a single continuous tone of 440Hz. If thespeed of sound in air is about 343m/s, then how far would we have to put the backspeaker behind the front one so that weneverhear the tone when standing anywheredirectly in front of the speakers?

Answers

Answer:

d= 0.3898 m

Explanation:

given,

frequency of the wave = 440 Hz

speed of the sound = 343 m/s

wavelength of the wave = ?

v = λ x f

[tex]\lambda = \dfrac{v}{f}[/tex]

[tex]\lambda = \dfrac{343}{440}[/tex]

λ = 0.7795 m

distance where he should be standing

if you line them up you will see the waves have cancelled each other out

if two speaker are lined together  

The speed of sound in the air has no relevance on this question as it would not matter how fast the waves traveled but only that they travel at the same speed as each other.

The distance of half a wavelength in this case is

d = λ/2

 d = 0.7795/2

d= 0.3898 m

Final answer:

To ensure silence in front of two speakers emitting a 440Hz tone due to destructive interference, the back speaker must be positioned an odd multiple of half the wavelength of the sound away from the front speaker, with the minimum distance being half the wavelength, 0.38975 meters.

Explanation:

To achieve silence in front of the speakers by exploiting destructive interference, the back speaker must be placed at a distance corresponding to an odd multiple of half the wavelength of the sound produced. Given that the sound has a frequency of 440Hz and the speed of sound in air is approximately 343m/s, we can calculate the wavelength using the formula \(\lambda = \frac{v}{f}\), where \(\lambda\) is the wavelength, \(v\) is the speed of sound, and \(f\) is the frequency. Substituting the given values, we find that the wavelength is \(\lambda = \frac{343 m/s}{440 Hz} = 0.7795 m\). To achieve destructive interference, the distance should be an odd multiple of half this wavelength, i.e., \((2n+1)\frac{\lambda}{2}\) where \(n\) is an integer starting from 0. Thus, the minimum distance required to never hear the tone directly in front of the speakers is \(0.7795 m / 2 = 0.38975 m\), which is half the wavelength.

A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C. What is the entropy change of the sample if the latent heat of vaporization of water is 2.26 x 10⁶ J/kg?

Answers

Answer:

The entropy change of the sample of water =  6.059 x 10³ J/K.mol

Explanation:

Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol

Mathematically, entropy is expressed as

ΔS = ΔH/T....................... Equation 1

Where ΔH = heat absorbed or evolved, T = absolute temperature.

Given:  If 1 mole of water = 0.0018 kg,

ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.

T = 100 °C = (100+273)  K = 373 K.

Substituting these values into equation 1,

ΔS =2.26x 10⁶/373

ΔS = 6.059 x 10³ J/K.mol

Therefore the entropy change of the sample of water =  6.059 x 10³ J/K.mol

The entropy change of the sample of steam is equal to 6,058.98 J/Kmol.

Given the following data:

Mass = 1.00 kgTemperature = 100°C to Kelvin = [tex]273+100=373\;K[/tex]Latent heat of vaporization of water = [tex]2.26 \times 10^6\;J/kg[/tex]

To determine the entropy change of the sample of steam:

Mathematically, entropy change is given by the formula:

[tex]\Delta S = \frac{\Delta H}{T}[/tex]

Where:

[tex]\Delta H[/tex] is the enthalpy change.T is the absolute temperature.

Substituting the given parameters into the formula, we have;

[tex]\Delta S = \frac{2.26 \times 10^6\times 1}{373}[/tex]

Entropy change = 6,058.98 J/Kmol.

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2. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are +7.0μC, -8.0 μC and -6.0 μC. Calculate the net force on charge 1 due to the other two charges in unit vector notation. Give values for the magnitude and direction of the force, too.

Answers

Answer:

0.53 N, 25.6°

Explanation:

side of triangle, a = 1.2 m

q = 7 μC

q1 = - 8 μC

q2 = - 6 μC

Let F1 be the force between q and q1

By using the coulomb's law

[tex]F_{1}=\frac{Kq_{1}q}{a^{2}}[/tex]

[tex]F_{1}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 8\times 10^{-6}}{1.2^{2}}[/tex]

F1 = 0.35 N

Let F2 be the force between q and q2

By using the coulomb's law

[tex]F_{2}=\frac{Kq_{2}q}{a^{2}}[/tex]

[tex]F_{2}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 6\times 10^{-6}}{1.2^{2}}[/tex]

F2 = 0.26 N

Write the forces in the vector form

[tex]\overrightarrow{F_{1}}=0.35\widehat{i}[/tex]

[tex]\overrightarrow{F_{2}}=0.26\left ( Cos60 \widehat{i}+Sin60\widehat{j}\right )[/tex]

[tex]\overrightarrow{F_{2}}=0.13 \widehat{i}+0.23\widehat{j}[/tex]

Net force

[tex]\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}[/tex]

[tex]\overrightarrow{F}=0.48 \widehat{i}+0.23\widehat{j}[/tex]

Magnitude of the force

[tex]F=\sqrt{0.48^{2}+0.23^{2}}[/tex]

F = 0.53 N

Direction of force with x axis

[tex]tan\theta =\frac{0.23}{0.48}[/tex]

θ = 25.6°

Final answer:

To calculate the net force on charge 1 due to the other two charges, we need to find the individual forces between charge 1 and the other charges and then combine them vectorially using Coulomb's law.

Explanation:

To calculate the net force on charge 1 due to the other two charges, we need to find the individual forces between charge 1 and the other charges and then combine them vectorially. The magnitude of the force between two charges can be calculated using Coulomb's law:

F = k * (|q1| * |q2|) / (r^2)

Where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.

Let's calculate the individual forces:

Force between charge 1 (+7.0μC) and charge 2 (-8.0 μC) can be calculated using the formula:
F12 = k * (|q1| * |q2|) / (r^2) = k * (|7.0| * |-8.0|) / (1.20^2)Force between charge 1 (+7.0μC) and charge 3 (-6.0 μC) can be calculated using the same formula:
F13 = k * (|q1| * |q3|) / (r^2) = k * (|7.0| * |-6.0|) / (1.20^2)

Next, we can calculate the net force on charge 1 by adding the forces vectorially:

Net Force on charge 1 = F12 + F13

The most powerful tugboats in the world are built in Finland. Theseboats exert a force with a magnitude of 2.85× 10^6N. Suppose one ofthese tugboats is trying to slow a huge barge that has a mass of 2.0× 10^7kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its maximum force for 21 s in the direction opposite to that in which thebarge is moving, what will be the change in the barge’s momentum? Howfar will the barge travel before it is brought to a stop?

Answers

Answer:

-5.985*10^7Ns=momentum

ds=31.578m

Explanation:

The most powerful tugboats in the world are built in Finland. Theseboats exert a force with a magnitude of 2.85× 10^6N. Suppose one ofthese tugboats is trying to slow a huge barge that has a mass of 2.0× 10^7kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its maximum force for 21 s in the direction opposite to that in which thebarge is moving, what will be the change in the barge’s momentum? Howfar will the barge travel before it is brought to a stop?

from newton's second law of motion , which states that the rate of change in momentum is directly proportional to the force applied

f=kdv/dt

k=1

ft=dv/dt

dp=dv/dt

f=negative because its moving in the opposite direction

2.85× 10^6*21=

-5.985*10^7Ns=momentum

-dp=mv2-mv1

-dp+mv1=mv2

-5.985*10^7+2.0× 10^7 kg*3.0 m/s=2.0× 10^7kgV2

v2=150000/(2.0× 10^7)

v2=0.0075m/s to the right

ds=1/2(v2+v1)dt

ds=0.5*(3.0075)*21

ds=31.578m

the barge will travel 31.578m before it is brought to rest.

Answer:

p = –5.985×10⁷ kg×(m/s)

x = 31.57875 m

The boat's change in momentum is –5.985×10⁷ kg×(m/s), and it will travel 31.57875 meters before it stops.

Explanation:

First off, let's take a quick look at the formula for momentum: momentum equals the force times the time the force is applied for. This means:

p = FΔt

Why is momentum represented by the letter p? That's actually a good question, to be honest—I guess they were running out of letters because the letter m is already used for "mass". But I digress.

Let's go back to the equation. We have both those values—the force applied and the time it is applied for—and so we can substitute those into the equation. But be careful: here, the force is applied in the opposite direction as the actual motion. This means the force must be negative. We get:

p = (–2.85×10⁶ N)(21 s)

p = –5.985×10⁷ kg·(m/s)

That's our value for initial momentum, not the final momentum. Why? Momentum and impulse are equivalent and their units are the same. If we calculated this as the impulse (J = mΔv), we'd be using the initial velocity, not the final velocity (the velocity after the force is applied).

But this problem asked us to solve for two things: the change in momentum (Δp) and the stopping distance (Δx). We can't find either value if we don't know the velocity of the block after the force is applied. This means we need to solve for it.

This is where knowing impulse and momentum are equal comes in handy: if the two values are equal, their formulas should be equal, too. It's like solving a system of equations with the substitution method. With the impulse and momentum formulas, we make them equivalent and get:

mΔv = FΔt

The change in something is like the difference in subtraction: it's the final value minus the initial value. We can then rewrite the equation as:

m(v₂ – v₁) = FΔt

Here, I used (v) for the final (second) velocity and (v) for the initial (first) velocity. Now, since we need to find the final velocity, (v), we need to isolate it by solving the equation for this value.

m(v₂ – v₁) = FΔt

[m(v₂ – v₁)] ÷ m = (FΔt) ÷ m

v₂ – v₁ = (FΔt) ÷ m

v₂ – v₁ + v₁ = [(FΔt) ÷ m] + v

v₂ = [(FΔt) ÷ m] + v

That's our formula to find the final velocity. The problem already gave us all the values we need to solve this equation: the force applied, the time the force is applied for, the mass of the object being stopped, and the object's initial velocity. When we substitute those into the equation, we get:

v₂ = [(FΔt) ÷ m] + v

v₂ = [(–2.85×10⁶ N)(21 s) ÷ 2.0×10⁷ kg] + 3.0 (m/s)

v₂ = [(–5.985×10⁷ kg·(m/s)) ÷ 2.0×10⁷ kg] + 3.0 (m/s)

v₂ = 3.0 (m/s) – 2.9925 (m/s)

v₂ = 0.0075 (m/s)

That gives us the final velocity that we need to find (Δp) and (Δx). Now, using the equations for each value, we can substitute in and finish this up!

Δp = mv) = m(v₂ – v₁)

Δp = (2.0×10⁷ kg)[0.0075 (m/s) – 3.0 (m/s)]

Δp = (2.0×10⁷ kg)[–2.9925 (m/s)]

Δp = –5.985×10⁷ kg·(m/s)

Δx = ¹/₂(v₂ + v₁)(Δt)

Δx = ¹/₂[0.0075 (m/s) + 3.0 (m/s)](21 s)

Δx = ¹/₂[3.0075 (m/s)](21 s)

Δx = [3.0075 (m/s)](10.5 s)

Δx = 3.157875 m

And there we go! Problem solved! I hope this helps you! Have a great day!

The current through a 10-ohm resistor connected to a 120-v power supply is

Answers

Answer:

12 A

Explanation:

Voltage, V = 120 V

Resistance, R  10 ohm

By using ohm's law

V = i x R

where, i is the current

i = V / R

i = 120 / 10

i = 12 A

thus, the current is 12 A.

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