An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the expansion is isothermal, what are the final pressure pfi and the work Wi done by the gas? If, instead, the expansion is adiabatic, what are the final pressure pfa and the work Wa done by the gas? State your answers in terms of the given variables.

Answers

Answer 1

Answer:

Isothermal :   P2 = ( P1V1 / V2 ) ,  work-done [tex]pdv = nRT * In( \frac{V2}{v1} )[/tex]

Adiabatic : : P2 = [tex]\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }[/tex]  , work-done =

W = [tex](3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)[/tex]

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

[tex]pdv = nRT * In( \frac{V2}{v1} )[/tex]

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

[tex]P1V1^y = P2V2^y[/tex]

where y = 5/3

hence : P2 = [tex]\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }[/tex]

Work-done

W = [tex](3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)[/tex]

Where    [tex]T2 = T1V1^(2/3)/V2^(2/3)[/tex]


Related Questions

You collect some data on horse racing along a straight track. You are able to fit the motion of the horse to a function during this interval, where you’ve chosen a particular spot on the track to be your origin and started your clock (t = 0) when you started collecting this new data.Required:a. What is the horse’s velocity as a function of time? Does the horse ever turn around during this time?b. What is its acceleration as a function of time?

Answers

Answer:

The equation is missing in the question. The equation is [tex]$10 m + 5(m/s^2)t^2+3(m/s^3)t^3$[/tex]

a). [tex]$v=10 t +9t^2$[/tex] , the horse will not turn.

b). a(t) = 10 + 18t

Explanation:

Given :

[tex]$x(t)=10 m + 5(m/s^2)t^2+3(m/s^3)t^3$[/tex]

∴ At t =0, x = 10 m

a). Velocity as a function of time

[tex]$v = \frac{dx}{dt} $[/tex]

  = [tex]$10 t +9t^2$[/tex]

Turning velocity must be zero.

v(t) = 0

[tex]$10 t +9t^2=0$[/tex]

[tex]$\therefore t = 0 \text{ or}\ t =-\frac{10}{9}$[/tex]

Taking the positive value of time.

The horse will not turn.

b). Acceleration as a function of time.

[tex]$a(t)=\frac{dv}{dt}$[/tex]

     = 10 + 18t

∴ a(t) = 10 + 18t

A crane raises a crate with a mass of 75 kg to a height of 10 m. Given that the
acceleration due to gravity is 9.8 m/s2, what is the crate's potential energy at
this point?
A. 720 J
B. 36,750 J
C. 7,350 J
D. 77 J

Answers

Answer: C. 7,350 J

Explanation:

Just did it in Apex

The crate's potential energy at this point will be 7,350 J.Option C is correct.

What is the potential of the potential energy?

The potential energy is the energy by the virtue of the position in the system. It varies with the height of the system. The potential energy unit is the joule (j).

The given data in the problem is;

m is the mass of the crate= 75 kg

h is the height = 10 m.

a is the acceleration due to gravity = 9.8 m/s²

P is the potential energy of the crate to be found as;

[tex]\rm PE = mgh \\\\ \rm PE =75 \times 9.81 \times 10 \\\\ PE=7350 \ J[/tex]

The potential energy at this point will be 7,350 J.

Hence option C is correct.

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Calculate the average speed of a complete round trip in which the outgoing 220 kmkm is covered at 92 km/hkm/h , followed by a 1.0-hh lunch break, and the return 220 kmkm is covered at 55 km/hkm/h . Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

60 km/hr

Explanation:

220 km is covered at 92km/hr in the round trip

There is 1 hour lunch break

During the return of the round trip 220 km is covered at 55km/hr

The first step is to calculate the total distance

= 220km + 220km

= 440km

The time spent during the entire round trip can be calculated as follows

= (220/92) + 1 + (220/55)

= 2.3913 + 1 + 4

= 7.3913

Therefore the average speed can be calculated as follows

= distance/ time

= 440/7.3913

= 59.52

= 60km/hr (approximated to 2 significant figures)

Hence the average speed is 60 km/hr

Convert a speed of 4.50 km/h to units of ft/min. (1.00 m = 3.28 ft).a. 165 ft/min.b. 82.3 ft/min.c. 886 ft/min.d. 246 ft/min.e. 0.246 ft/min.

Answers

Answer:

246ft/min

Explanation:

In this problem we are faced with unit conversion.

We are expected to convert from km/h to ft/min

Given that

1.00 m = 3.28 ft

1km(1000m)= 3280 ft

But we need to convert 4.5km to ft

Hence in ft 4.5km= 3280*4.5

= 14760ft

We now convert hour to minutes

60min make 1 hour

Therefore the convert form km/h to ft/min is

= 14760/60= 246ft/min

The answer is 246ft/min

What statements are true regarding the skydivers jump and fall to Earth?
Regardless of their weight, before opening their parachutes, the guys felt at the same velocity.
According to the data, the acceleration due to gravity is 9.8m/s/s.
Once the parachutes were opened, all five skydivers continued to fall at exactly the same velocity
Opening the parachutes provided air resistance and changed their velocity
By free-falling for a longer period of time, the skydivers would have changed the acceleration due to
gravity

Answers

Answer: Regardless of their weight, before opening their parachutes, the guys fell at the same velocity.

According to the data, the acceleration due to gravity is 9.8 m/s/s

Opening the parachutes provided air resistance and changed their velocity.

Explanation:

When in free fall, mass does not affect the speed of moving objects (only gravitational force is acting upon it).

If you take a look at the data, you can see that the velocity is increasing at a constant rate in intervals of 9.8. This is the acceleration which makes sense because acceleration is the change of velocity divided by time.

Think about it: parachutes slow down the object that is falling because it creates a barrier between the object falling and the gravitational force it encounters. The change in the effects of the gravitational force will alter the velocity.

I hope this was helpful. Have a nice day!

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air resistance)?

Answers

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

[tex]t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s[/tex]

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

[tex]d=ut-\dfrac{1}{2}gt^2[/tex]

Putting all values

[tex]d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m[/tex]

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

A hot air balloon is ascending straight up at a constant speed of 6.60 m/s. When the balloon is 11.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 30.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground level are these places

Answers

Answer:

[tex]H_1 =39.05 \ m[/tex] OR [tex]H_2 =14.5 \ m)[/tex]

Explanation:

From the question we are told that

The constant speed of the balloon is [tex]v = 6.60 \ m/s[/tex]

The height of the balloon is [tex]h = 11.0 \ m[/tex]

The initial speed of the pellet is [tex]u = 30 \ m/s[/tex]

Generally the height of the balloon at the point it is the same altitude with the pellet is mathematically represented as

[tex]H = h + v * (t)[/tex]

Note: vt is the distance covered by the balloon before the pellet got to it

Generally the height of the pellet when it is the same height with the balloon is mathematically represented using kinematics equation

[tex]s = ut + \frac{1}{2} gt^2[/tex]

So

H = s

=> [tex] ut + \frac{1}{2} gt^2 = h + v * (t)[/tex]

=> [tex] 30t + \frac{1}{2} *( -9.8)t^2 = 11 + 6.60t[/tex]

=> [tex] 4.9t^2 -23.4t + 11= 0[/tex]

using the quadratic formula to solve the above equation

From the quadratic formula calculation

[tex]t_1 = 4.25 \ s[/tex]

   OR

     [tex]t_1 =  0.529  \  s[/tex]

So the height of this two place above the ground is mathematically evaluated as

[tex]H_1 = h + v * (4.25)[/tex]

[tex]H_1 = 11 + 6.60 * (4.25)[/tex]

[tex]H_1 =39.05 \ m[/tex]

OR

[tex]H_2 = h + v * ( 0.529)[/tex]

[tex]H_2 = 11 + 6.60 * (0.529)[/tex]

       [tex]H_2  =14.5 \  m)[/tex]

The speed of sound in humid air than in dry air. Why ​

Answers

Answer:

When moisture is removed from air, its density increases. The speed of sound in a humid air is inversely proportional to the square root of its density. Therefore, the speed of sound in moist air is humidity increases, the velocity of sound increase and vice-versa.

Hope it helps :)

To understand the electric force between charged and uncharged conductors and insulators. When a test charge is brought near a charged object, we know from Coulomb's law that it will experience a net force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.Consider three plastic balls (A, B, and C), each carrying a uniformly distributed charge equal to either +Q, -Q or zero, and an uncharged copper ball (D). A positive test charge (T) experiences the forces. The test charge T is strongly attracted to A, strongly repelled from B, weakly attracted to C, and strongly attracted to D. Assume throughout this problem that the balls are brought very close together. What is the nature of the force between balls A and B? a. Strongly attractive b. Strongly repulsive c. Weakly attractive d. Neither attractive nor repulsive

Answers

Answer:

the correct  is a  Strongly ATTRACTIVE

Explanation:

For this exercise we must use that charges of the same sign repel and charges of the opposite sign attract, the attraction is strong if they are charged or weak if the charges are induced.

Let's apply this to our case.

The test load T is attracted by the sphere A, this implies that the charges are of different sign

the test charge T is repelled by the sphere B, therefore the charges are of equal sign

As the test charge cannot change the sign, this implies that the spheres A and B are of different sign, therefore attractive forces.

Now let's analyze the intensity, as in the exercise they indicate that spheres A and B are charged and are insulators, these charges cannot move, so the attraction must be Strong.

When reviewing the statements, the correct one is a  Strongly ATTRACTIVE

The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevelled A over m squared close parentheses. The wire conductivity is 2 cross times 10 to the power of 7 space open parentheses bevelled S over m close parentheses. Find the voltage drop across the wire. (Answer with the numeric value, don't write the unit V)

Answers

Answer:

The voltage drop across the wire is 2 V

Explanation:

Given;

length of wire, L = 10 m

current density, I/A, μ = 4 x 10⁶ (A/m²)

wire conductivity, σ = 2 x 10⁷ (S/m)

The resistivity of wire is given by;

[tex]\rho = \frac{RA}{L} \\\\But \ R = V/I\\\\\rho = \frac{VA}{IL}[/tex]

Conductivity, σ = ¹/ρ

[tex]\sigma = \frac{IL}{VA}\\ \\V = \frac{IL}{ A \sigma}\\\\V = (\frac{I}{A})\frac{L}{\sigma}\\ \\V = (\mu)\frac{L}{\sigma}\\\\V = (4*10^{6} )*\frac{10}{2*10^{7} } \\\\V = 2 \ V[/tex]

Therefore, the voltage drop across the wire is 2 V

An object travels due North and covers 500 miles in 5 hours. Which one of the following is the average velocity of the object?
a) 100 mph
b) 2500 mph
c) 2500 mph due North
d) 100 mph due North

Answers

Answer:

100 mph.

Explanation:

Given that,

Distance covered by an object is 500 miles in 5 hours.

We need to find the average velocity of the object. It can be given by :

[tex]v=\dfrac{d}{t}\\\\v=\dfrac{500\ \text{miles}}{5\ \text{hour}}\\\\v=100\ \text{mph}[/tex]

Hence, the average velocity of the object is 100 mph.

Why does wine go sour faster
it the cork is removed from the
bottle

Answers

When the cork is removed the gases from the wine are reacting to the oxygen and it creates a chemical reaction while when the cork is still on the chemicals never interact and stay there in the bottle



A basketball is shot. After the ball leaves the player's hand, in which direction does the ball accelerate?
A)It always accelerates in the opposite direction that the object is moving.
B) It always accelerates in a upward direction.
C) It always accelerates in a downward direction.
D) It accelerates upward when the ball is rising and downward when it is falling.

Answers

Answer: Always accelerates in a downward direction

Explanation: Gravity is pulling the ball downward

A 50.0-g Super Ball traveling at 25.5 m/s bounces off a brick wall and rebounds at 21.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.45 ms, what is the magnitude of the average acceleration of the ball during this time interval

Answers

Answer:

The magnitude of the average acceleration of the ball during this time interval is 10,449.44 m/s²

Explanation:

Given;

mass of super ball, m = 50 g

initial velocity of the ball, u = 25.5 m/s

final velocity of the ball, v = -21 m/s (re-bouncing backward)

time in contact with the wall, t = 4.45 ms = 0.00445 s

The average acceleration of the ball during this time interval is given by

[tex]a = \frac{dv}{dt} = \frac{v-u}{t} \\\\a = \frac{-21-25.5}{0.00445}\\\\a = -10449.44 \ m/s^2\\\\|a| = 10,449.44 \ m/s^2[/tex]

Therefore, the magnitude of the average acceleration of the ball during this time interval is 10,449.44 m/s²

The amount of kinetic energy an object has depends on which feature of the object?

Answers

Answer:

Mass and speed.

Explanation:

The amount of kinetic energy of an object depends on the object's mass and speed.

Answer:

its a motion hope this helps

Explanation:

1 gram is 0.035 ounces.How many ounces is 200 grams

Answers

Answer:

7.05479

Explanation:

Answer:

200 grams:     ounces:7.05479239 that's the answer

Explanation:

Which statement describes a controlled experiment?

A. It includes more than one control group.

B. It includes a control group and an experiment group.

C. It includes only the control group.

D. It includes only an experimental group.

Answers

Answer:

B

Explanation:

I think this is right.

If it is Good luck!

The controlled experiment includes a control group and an experiment group, this statement describes a controlled experiment, therefore the correct answer is option B.

What is the scientific investigation?

Scientific discovery is the method of finding solutions through in-depth analysis and discovering them through the outcomes of experiments.

True experimental outcomes that can be backed up by data are incredibly important to scientific inquiry.

A regulated investigation where all sensors to measure have been held constant and the scientific element is utilized as a factor associated.

Since the statement depicts a controlled experiment with a control group and an experiment group, option B is the appropriate response.

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how was the atomic bomb different from conventional bombs? ​

Answers

Answer:

A conventional bomb releases most of its energy in the form of blast. Atomic bombs on the other hand, release 50 per cent energy as blast, 35 per cent as heat and 15 per cent as nuclear radiation.

Explanation:

A conventional bomb releases most of its energy in the form of blast. Atomic bombs on the other hand, release 50 per cent energy as blast, 35 per cent as heat and 15 per cent as nuclear radiation.

A mountain lion jumps to a height of 3.25 m when leaving the ground at an angle of 43.2°. What is its initial speed (in m/s) as it leaves the ground?

Answers

Recall that

[tex]{v_f}^2={v_i}^2+2a\Delta y[/tex]

where [tex]v_i[/tex] and [tex]v_f[/tex] are the lion's initial and final vertical velocities, [tex]a[/tex] is its acceleration, and [tex]\Delta y[/tex] is the vertical displacement.

At its maximum height, the lion has 0 vertical velocity, so we have

[tex]0={v_i}^2-2gy_{\rm max}[/tex]

where g is the acceleration due to gravity, 9.80 m/s², and we take the starting position of the lion on the ground to be the origin so that [tex]\Delta y=y_{\rm max}-0=y_{\rm max}[/tex].

Let v denote the initial speed of the jump. Then

[tex]v_i=v\sin(43.2^\circ)=\sqrt{2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(3.25\,\mathrm m)}\implies\boxed{v\approx11.7\dfrac{\rm m}{\rm s}}[/tex]

The initial speed will be "11.7 m/s".

Speed and Displacement:

Given:

Height = 3.25 mAngle = 43.2°Acceleration due to gravity = 9.8 m/s²

We know,

→ [tex]vf^2=v_i^2+2a\Delta y[/tex]

At max. height, vertical height be zero, then

→ [tex]0 = v_i^2-2gy_{max}[/tex]

or,

→ [tex]\Delta y = y_{max} -0 = y_{max}[/tex]

now,

The initial speed,

→ [tex]v_i = v \ Sin(43.2^{\circ})[/tex]

By substituting the values,

      [tex]= \sqrt{1\times 9.8\times 3.25}[/tex]

      [tex]= 11.7 \ m/s[/tex]

Thus the solution above is correct.

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What positively charged subatomic particle is found in the nucleus?

Answers

Answer:

Protons

Explanation:

The Proton is one of the subatomic particles that are found in the nucleus of an atom. The protons are positively charged. They have an electric charge of 1.6 * 10^-19 C. Protons have an approximately 1 atomic mass unit, or otherwise called, 1 amu.

One or more protons, at least, are found in every nucleus if an atom, as they are an essential and very necessary part of the nucleus. Ernest Rutherford, in 1920, named protons, proton

A computer monitor uses 200 W of power. How much energy does it use in
10 seconds?
A. 20 J
B. 200 J
C. 20,000 J
D. 2000 J
SUBMIT

Answers

Answer: D (2000 J)

Explanation: p= 200 W

t= 10 s

E=pt

E=200 W * 10s

E= 2000 J

Answer:

it is D.

Explanation:

Suppose you figured resistance of a color-coded resistor is 100 ± 5 ohms. You did the Ohms law experiment and measured resistance of the same resistor. You found the measured value is 92 ± 4 ohms. How would you compare the two values? Choose all that are correct. Group of answer choices I will find the percent difference between 100 and 92 ohsm and conclude that they are different I will look at the range of both, such as 100-5 = 95 and 94+4=98. Since the numbers are different, I conclude that they are different within uncertainty I will look at the range of both, such as 100-5 = 95 and 94+4=98. Since the ranges overlap, I can conclude that they are the same within uncertainty. I will draw a dot and whisker plot for both. Since they overlap, I can conclude that they are the same within uncertaitny

Answers

Answer:

* I will look at the range of both, such as 100-5 = 95 and 94+4=98. Since the ranges overlap, I can conclude that they are the same within uncertainty.

* I willl draw a dot-and-whisker plot for both. Since they overlap, I can conclude that they are equal within the uncertainty

Explanation:

The value of all physical measurement has an uncertainty, so it is within a range of values, in this case we must compare the range of values ​​of the resistors

The nominal value is 100 ± 5 Ω

therefore its range of possible values ​​is between

             95 <= R <= 105

For the experimentally measured value

          88 <= R <= 96

as the values ​​overlap we must conclude that they are the same.

When reviewing the different options we have two statements that are correct

* I will look at the range of both, such as 100-5 = 95 and 94+4=98. Since the ranges overlap, I can conclude that they are the same within uncertainty.

* I willl draw a dot-and-whisker plot for both. Since they overlap, I can conclude that they are equal within the uncertainty

Which laboratory activity involves a chemical change?

Answers

Answer:

A: leaving a copper penny in vinegar until it turns green

Explanation:

The corrosion is from oxidation -- a chemical reaction between the metal and oxygen, water, and carbon dioxide in the air. Rust is the term used to describe this process when it happens to iron instead of other metals. I also took it on E2020. If you need further explanation, please comment below.

The laboratory activity which involves a chemical change is leaving a copper penny in vinegar until it turns green.

What are chemicals?

The chemicals are the artificially prepared or purified and a distinct compound or substance.

The corrosion is the result of oxidation that is a chemical reaction between the metal and oxygen, water, and carbon dioxide in the air. Rust is the term used to describe the corrosion process.

Corrosion also takes place in copper due to which the copper turns green when oxidized.

Thus, The laboratory activity which involves a chemical change is leaving a copper penny in vinegar until it turns green.

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Which of these statements describes a species? Group of answer choices
A. A species is a group of organisms that are closely related and produce fertile offspring
B. A species is a group of organisms that are closely related.
C. A species is a group of organisms that live in the same habitat
D. A species is a group of organisms that live in an area and interact.

Answers

I believe the answer would be A .
The answer haves to be A or it could be also D but most likely I am positive that is A

Question 2
What two qualities of objects does gravity depend on?
What each object is made up of and the distance between them.
What each object is made up of and what material is between them.
Each object's mass and its shape.
Each object's mass and the distance between them.

Answers

Answer:

in pretty sure the last one

The qualities of objects do gravity depend on Each object's mass and the distance between them. Hence the correct option is (4).

Gravity is the force of attraction between two objects with mass. The strength of the gravitational force depends on the mass of each object and the distance between them.

The mass of an object refers to the amount of matter it contains. The greater the mass of an object, the stronger its gravitational pull.

The distance between two objects is also a crucial factor. The gravitational force weakens as the distance between the objects increases. In other words, the force of gravity decreases with increasing distance.

So, the two qualities that gravity depends on are each object's mass and the distance between them.

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Please HELP I NEED TO WRITE A PARAGRAPH BUT DON'T KNOW HOW TO EXPLAIN IT!

How does protein, atoms, vitamin d, magnesium, and zinc work together?

Answers

Explain how they interact with each other. I'm not a chem genius but taht would make the most sense.

Hope it helps! Comment if you have any questions and have an amazing day!

Newtons first law of motion describes

Answers

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

A person running down the hallway at a velocity of 4.2 m/s comes to a stop in a time of 1.8 seconds to avoid hitting the wall. What was the acceleration of the person?
A.) 7.6 m/s/s
B.) -7.6 m/s/s
C.) 2.3 m/s/s
D.) -2.3 m/s/s

Answers

Hshshshhshdhdhdhdhddhdhdhdhdhdh

Answer:

It's not -7.6

Explanation:

All I can tell you

For a point charge, how does the potential vary with distance from the point charge, r?
b. r.c. 1/r.d. 1/r2.e. r2.

Answers

For a point charge, how does the potential vary with distance from the point charge, r?

a constant

b. r.

c. 1/r.

d. [tex]1/r^2[/tex].

e. [tex]r^2[/tex].

Answer:

The  correct option is  C

Explanation:

Generally for a point charge the electric potential is mathematically represented as

    [tex]V  =  \frac{k  Q  }{r }[/tex]

Here we can deduce that the electric potential varies inversely with the distance i.e

      [tex]V  \  \alpha \  \frac{1}{r}[/tex]

So

   

The potential should vary with distance from the point charge r should be option c. 1/r.

What is a point charge?

The point charge with respect to the electric potential should be expressed in

V = KQ/r

Here we can say that the electric potential should be varied oppositely with the distance

Here is the electric file with respect to the point charge should be obtained from the Coulomb law. It should be radially outward from the point charge in all types of directions.

Therefore, the option c is correct.

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Suppose you are an astronaut and you have been stationed on a distant planet. You would like to find the acceleration due to the gravitational force on this planet so you devise an experiment. You throw a rock up in the air with an initial velocity of 9 m/s and use a stopwatch to record the time it takes to hit the ground. If it takes 6.4 s for the rock to return to the same location from which it was released, what is the acceleration due to gravity on the planet?

Answers

Answer:

1.40625m/s²

Explanation:

Using the equation of motion expressed as v = u+gt where;

v is the  final velocity of the ball

u is the initial velocity

g is the acceleration due to gravity

t is the time taken

Given

u = 9m/s

v = 0m/s

t = 6.4s

Required

acceleration due to gravity g

Since the rock is thrown up, g will be a negative value.

v = u+(-g)t

0 = 9-6.4g

-9 = -6.4g

6.4g = 9

divide both sides by 6.4

6.4g/6.4 = 9/6.4

g = 1.40625m/s²

Hence the acceleration due to gravity on the planet is  1.40625m/s²

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