Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.

Answers

Answer 1

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The expression is  [tex]W_c =  P_o V_o ln (R_v)[/tex]  

Explanation:

Generally smallest workdone done by  a gas is mathematically represented as

          [tex]dW  =  PdV[/tex]

Generally for an isothermal process

    [tex]PV  =  nRT = constant [/tex]

=>   [tex]P = \frac{nRT}{V}[/tex]

Generally the total workdone is mathematically represented as

   [tex]W_c =  \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]

=> [tex]W_c = nRT  \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]

=>  [tex]nRT [lnV]   | \left \ {V_f}} \atop {V_o}} \right.[/tex]

=>  [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]

=>  [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]

From the question [tex]\frac{V_f}{V_o }  =  R_v[/tex]

=> [tex]W_c =  P Vln (R_v)[/tex]

at initial  state

[tex]W_c =  P_o V_o ln (R_v)[/tex]  

Calculate The Work WC Done By The Gas During The Isothermal Expansion. Express WC In Terms Of P0, V0,
Calculate The Work WC Done By The Gas During The Isothermal Expansion. Express WC In Terms Of P0, V0,

Related Questions

Silly Goose falls 1.0 m to the floor. How long does the fall take

Answers

Answer:You need to give more explanation sorry

Explanation:

Answer:

4.20 seconds

Explanation:

Supposing that silly goose weighs 69 pounds, we need to start on the math.

Simple maths, truly and really. 69/1=69, of course.

Therefore it will take 4.20 seconds for silly goose to hit the ground. if he is going to be a silly goose though, he can just go in the pond, instead of wasting his time.

The boys are finally old enough to compete in the box car derby race at the local fair. They have been working on their cars since the conclusion of the race last year. One boy's car raced down the track and placed 2nd in his race. However, the other boy's car started well but half-way through the race a wheel came off and his car came to a complete stop. The boy was very disappointed and the other boy felt horrible for his friend. Which of the following graphs best represents the motion of boy's car that stopped?

Answers

i think it’s b because they both stumbled up a hole

A soccer ball accelerates from rest and rolls 6.5m down a hill in 3.1 s. It then bumps into a tree. What is the speed of the ball just before it hits the tree.

Answers

Answer:

2.096m/s

Explanation:

The speed of this soccer ball can be calculated using the formula;

Speed = distance/time

According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;

Speed = 6.5/3.1

Speed of the ball = 2.096m/s

Therefore, the speed of the ball before hitting the tree is 2.096m/s

A recipe gives the instructions below
After browning the meat pour off fat from the pan to further reduce fat use a strainer.

what type lf separation methods are described in the recipe

A decantation and screening
B distillation and screening
C decantation and centrifugation
D distillation and filtration

Answers

Answer:

A. decantation and screening

Explanation:

Decantation is the one of the process of separating the mixture. In this process the precipitated liquid is separated from the solid. According to the given instruction for the recipe, the fat which is in liquid state is separated from meat. In the process of screening, more liquid is separated by placing the mixture on the screen. Here, the gravity plays an important role for the process of separation.  

Answer:

a

Explanation:

Please provide explanation!!!
Thank you.

Answers

Answer:

(a) 102 cm/s

(b) 0.490 cm²

Explanation:

(a) Use Bernoulli equation.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0

½ ρ v₁² + ρgh₁ = ½ ρ v₂²

½ v₁² + gh₁ = ½ v₂²

½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²

v = 102 cm/s

(b) The flow rate is constant.

v₁ A₁ = v₂ A₂

(25.0 cm/s) (2.00 cm²) = (102 cm/s) A

A = 0.490 cm²

If 0.5 kg of this material is used in a transformer core, how long would it have to operate at a frequency of 60 cps to heat up 1

Answers

Answer:

Hello your question is incomplete attached below is the complete question

answer : 49 seconds

Explanation:

considering only Hysteresis loss

we have to calculate the Area affected/under the  Hysteresis loss

= volume * area

= 4 * ( 1.5 * 20 ) =  120 tesla. A/m

next we calculate the volume of the material

= mass of material / density

= 500 grams / 7.9 g/cm^3  = 6.33 * 10^-5 m^3

next we calculate the heat lost per cycle

= 6.33 * 10^-5 m^3 * 120 = 0.007596 joules

The total heat required to raise  temperature by 1°c

= Cp * ΔT * n

= 3R * n * ΔT  = 3(8.314) * 8.95 * 1  = 223.23 Joules

where n = number of moles = 500grams / 55.85 = 8.95moles

ΔT = 1

Therefore the time required to have to operate at a frequency of 60 cps

= Total heat required  / heat lost per cycle

=( 223.23 / 0.007596 ) 60 cps

= 489.796 seconds ≈  49 seconds

A dog has a mass of 60kg and an acceleration of 2m/s/s. What is the force of the dog?

Answers

The force 120 Newton’s

It took 50 joules to push a chair 5 meters across the floor. With what force was the chair pushed?

Answers

Answer:

The chair was pushed with 10 N.

Explanation:

The chair was pushed with 50 Joules.  

Work = Force * Distance

50 J = F * 5m  

F = 50 / 5 = 10N  

The chair was pushed with 10 N.

The chair was pushed with 10 N force.

What is Work done?

Work is defined as the measure of energy transfer that occurs when an object is moved over a distance by an external force, at least part of which is applied in the direction of displacement.

If the force is constant then work can be calculated by multiplying the length of the path by the component of the force acting along the path, which is expressed mathematically as work W equal to the force f over a distance d, or W = fd.

So, for above given information,

Work done= 50 joules

Distance covered by the chair = 5m

Then, Force= W/d

=50/5= 10N

Thus, the chair was pushed with 10 N force.

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How do I proton and and electron compared

Answers

What’s the question?

HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the
motion of the ball?

Answers

Answer:

Option C.

Explanation:

To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:

1. Determination of the velocity.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Final velocity (v) =.?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 100)

v² = 0 + 1960

v² = 1960

Take the square root of both side.

v = √(1960)

v = 44.27 m/s

2. Determination of the time taken.

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Time (t) =.?

h = ½gt²

100 = ½ × 9.8 × t²

100 = 4.9 × t²

Divide both side by 4.9

t² = 100 / 4.9

Take the square root of both side

t = √(100 / 4.9)

t = 4.52 s

From the above illustration,

Initial time (t1) = 0 s

Final time (t2) = 4.52 s

Initial velocity (u) = 0 m/s

Final velocity (v) = 44.27 m/s

Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.

A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east. what is the distance and displacement

Answers

Explanation:

Let east = E, and, west = opposite to east = - E.

Here, displacement:

=> 2m east + 4m west + 1m east

=> 2E + 4(-E) + 1E

=> 2E - 4E + 1E

=> - 1E

=> 1(-E)

=> 1m west

And, distance,

=> 2m + 4m + 1m = 7m

The distance of a person is 7 m and the displacement of the person is 1m west.

To find the distance and displacement, the given values are,

A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.

What is the distance and the displacement?

Displacement:

The displacement is shortest distance between initial and final position or we can say it is the straight line distance between initial and final position.If object moves in a straight line path without any turn then the path length and the displacement is always same.

Distance:

The distance is the total path length of the object while it will move from initial to final position.If the object move on curved path then displacement is smaller than the distance moved by the object.

Let us consider East = E and west = opposite to east = - E.

Calculating the displacement:

= 2m east + 4m west + 1m east

= 2E + 4(-E) + 1E

= 2E - 4E + 1E

= - 1E

= 1(-E)

= 1m west.

The displacement is 1m west.

Now calculating the distance,

= 2m + 4m + 1m

= 7m

The distance is 7m.

Thus, the displacement and the distance is found as 1 m west and 7m.

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Examine the diagram below.
Which of the above objects is moving the fastest?
A.
All 3 objects are moving at the same speed.
B. B
C. C
D. A

Answers

Answer:

Correct choice: D. Object A is the fastest

Explanation:

In a distance vs time graph, the distance is plotted vertically, and the time is plotted horizontally.

The diagram shows three graphs of objects A, B, and C.

The graph of A shows the object traveled 12 meters in 3 seconds, for a speed of 12/3= 4 m/s.

The graph of B shows the object traveled 8 meters in 4 seconds for a speed of 8/4=2 m/s.

Finally, the object C travels 4 meters in 4 seconds, for a speed of 4/4= 1 m/s

Thus, the fastest object is A.

when hydrogen shares electrons with oxygen the outermost shell of the hydrogen atoms are full with how many electrons? and oxygens valence shell is full with how many electrons? because the valence shells of these atoms are full,the atoms are stable.​

Answers

Answer:

2 and 8

Explanation:

please mark me brainiest I would really appreciate it.

How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J

Answers

Answer:

B) 200 [J]

Explanation:

In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]

F = m*g

where:

m = mass = 5 [kg]

g = gravity acceleration = 10 [m/s^2]

F = 5*10 = 50 [N]

w = F*d

where:

F = force = 50 [N]

d = 4 [m]

w = 50*4 = 200 [J]

One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.

Answers

Answer:

The value  is  [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

Explanation:

From the question we are told that  

   The speed of the airplane from the north is [tex]\frac{dN}{dt}  =  -181 \  km /hr[/tex]

The negative sign is because the direction is towards the south

  The speed of the airplane from the east is  [tex]\frac{dE}{dt}  =  -278 \  km/hr[/tex]

The negative sign is because the direction is towards the west

   The distance of the southbound plane from the airport is  [tex]N  =  30 \  km[/tex]

   The distance of the westbound plane is  [tex]E =  15 \  km[/tex]

Generally the distance between the plane is mathematically represented using Pythagoras theorem  as

    [tex]R^2  = N^2 + E^2[/tex]

Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes

So

     [tex]2R\frac{dR}{dt} =  2N \frac{dN}{dt} +   2E\frac{dE}{dt}[/tex]

Here

     [tex]R = \sqrt{N^2 + E^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R =33.54 \ m [/tex]

    [tex]2(33.54) * \frac{dR}{dt} =  2( 30)*(-181)  +   2*15*(-278)[/tex]

=>   [tex] 67.08 * \frac{dR}{dt} =  -19200[/tex]

=>   [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

The rate of change of the distance between the planes is 286.23 km/hr.

The given parameters;

speed of the airplane from North, dn/dt = 181 Km/hspeed of the airplane from the East, de/dt = 278 km/hnorth distance, n = 30 kmeast distance, e= 15 km

The distance between the two planes is calculated by applying Pythagoras theorem as shown below;

[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]

The rate of change of the distance between the planes is calculated as follows;

[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]

Thus, the rate of change of the distance between the planes is 286.23 km/hr.

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1. What is Ohm"s law?
2. If you placed a negatively charged hairbrush near your hair, what charge would your hair be?
3. You must change a lightbulb and the new lightbulb has a larger resistance. If the voltage of the battery does not change, what happens to the current going through the flashlight?
HELLPPPP

Answers

1. Ohm's law shows the relationship between:

voltagecurrentresistance

Formula: voltage = current x resistance

2. The negative charge on the hairbrush will induce a positive charge on your hair. As a result, your hair is going to be attracted to the hairbrush (and repelled by other strands of hair.)

3. V = IR, so if the resistance of the current increases, and the voltage of the current stays the same, there is as a result, going to be less current.

Best of Regards!

Research has shown that this type of interview is the most effective in predicting later job
performance.

Answers

Answer:

Situational Interview

Explanation:

A situational interview is about as close to the real job as it gets. During this type of interview, candidates may be presented with a visual or audio simulation of a scenario and asked to respond to it. They are asked to analyze a problem and profer suggestions on how they would handle it.

If the candidate has solved similar problems in the past, it will come to the fore.

If they haven't then the best outcome is that it will tell the interviewers how well the candidate is able to solve similar problems.

An example of a Situational Interview question is this:

How would you handle an angry customer who for no justifiable reason has decided to create a problematic scene to disrupt the business?

Because Situational Interviews are about behavioral responses (present, past, and future), they are powerful tools in determining or predicting future job performance. An interviewing technique that is developed using this methodology is called the S.T.A.R.

This is an acronym for Situation, Task, Action, Result.

Situation: the candidate is asked to present a challenging situation that occurred recently. This tests what the candidate sees as a challenging situation.

Task: The candidate based on the situation is asked to identify what they need to do to remedy the problem. This tells the interviewer(s) whether or not the candidate can think up a solution for the problem.

Action: Here they define the actual steps taken to resolve the problem

Result: The candidate against the above is required to give the result gotten

Action and Result tell the interviewer the quality of the candidate's ability to follow through and achieve the intended results. This also judges the quality of execution in terms of cost and time. The candidate with the lowest cost and time and the highest quality of outcome is considered the best.

Cheers

What is the key for a successful relationship? and Why?

Answers

Answer:

communication, if you don't talk you'll never know what's going on.

Explanation:

Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion is reversible but not isothermal and that the temperature decreases as the expansion proceeds. (a) Find an expression

Answers

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas  = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 240 K

= initial volume of gas  =  

= final volume of gas  =  

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

what is the meaning of the word physics​

Answers

Answer:

the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

Explanation:

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The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm

Answers

Answer:

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Explanation:

The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:

[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]

[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.

[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.

And we expand the equation above by definitions of elastic potential energy and kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]

[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the block, measured in kilograms.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.

[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.

If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:

[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]

[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

A
6. All other changeable factors that must
be kept the same to ensure a fair test
(what you keep the same).​

Answers

Answer:

a constant variable?

Explanation:

A constant variable is any aspect of an experiment that a researcher intentionally keeps unchanged throughout an experiment.

Experiments are always testing for measurable change, which is the dependent variable. You can also think of a dependent variable as the result obtained from an experiment. It is dependent on the change that occurs

waht is science
wjwissbsskdldmndndnd​

Answers

Answer:

the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.

Explanation:

A .05 kg rubber ball is dropped and hits the floor with an initial velocity of 10 m/s. It rebounds away from the floor with a final speed of 7 m/s after being in contact with the floor for .01 seconds. Find the magnitude of the force exerted by the floor on the rubber ball.

Answers

Answer:the answer is 3

Explanation:

Grass and plants get energy from
А
the sun.
B
eating food.
с
windmills.
D
electrons.

Answers

A is the answer for that question

Answer:

From the Sun

Explanation:

Plants can't eat any food. They don't ue or need windmills to get energy. They are plants so they don't have any electrons. The only way that they can recive energy from is the sun. Sometimes plants die when they don't get enough sun because they don't have any energy to live.

A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person

Answers

Answer:

20000

Explanation:

Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000

A designer is creating an obstacle for an obstacle course where a person starts on a moveable platform of height H from the ground. The person grabs a rope of length L and swings downward. At the instant the rope is vertical, the person lets go of the rope and attempts to reach the far side of a water-filled moat. The left side of the moat is directly below the position where the person will let go of the rope. The designer runs several tests in which the rope has different lengths and moves the platform such that the rope is always initially horizontal. The designer notices that the person cannot land on the other side if the length L is very short. The designer also notices that the person also cannot land on the other side if the length L is very close to the height H.

Assume the size of the person is much smaller than the lengths L and H. Let D represent the horizontal distance from below the release point to where the person lands.

Required:
a. Why does the person land in the moat if the rope's length is very short?
b. Why does the person land in the moat if the length is nearly the same as the height of the platform?

Answers

Answer:

* when L → H    chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

         the speed of the platform is very small, so we do not have the minimum required value

        vox = √ (g / (2 (H)) D

Explanation:

For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.

Let's start with the projectile launch

as the body leaves the vertical its velocity must be horizontal

         x = v₀ₓ t

         y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero

         t = √ 2 y₀ / g

we substitute

        x = vox √2y₀ / g

        v₀ₓ = √(g / 2y₀)     x

In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)

       v₀ₓ = √(g /(2 (H -L))    D

this is the minimum speed to cross the well.

Now let's use conservation of energy

starting point. On the platform

      [tex]Em_{o}[/tex] = U = m g H

final point. At the bottom of the swing

      Em_{f} = K + U = 1 / 2m v² + m g (H -L)

as there is no friction the mechanical energy is conserved

        Em_{o} = Em_{f}

       m g H = 1 / 2m v² + m g (H -L)

        v = √ (2gL)

let's write our two equations

the minimum speed to cross the well

       v₀ₓ = √ (g /(2 (H -L))    D

the speed at the bottom of the oscillatory motion

       v = √ (2g L)

we analyze the extreme cases

* when L → H    chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

         the speed of the platform is very small, so we do not have the minimum required value

        vox = √ (g / (2 (H)) D

From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well

      V₀ₓ = v

      g / (2 (H -L) D² = 2g L

       4 L (H- L) = D²

        4 H L - 4 L2 - D² = 0

        L² - H L - D² / 4 = 0

let's solve the quadratic equation

      L = [H ± √ (H2-D2)] / 2

we assume that H> D

       L = ½ H [1 + - RA (1 - (D / H) 2)]

The two values ​​of La give the range of values ​​for which the two speeds are equal

A) The person lands in the moat if the rope's length is very short because :

The speed of the platform is less than the required minimum speed

B) The person lands in the moat if the rope length is similar to the height of the platform because :

The speed required to cross the moat approaches infinity

Following the assumptions;

size of the person is much smaller than L and H

D = horizontal distance

The conditions that will cause the person to land on the moatThe person will land in the moat when the rope's length is very short because as the rope reduces in length the speed reduces as well such that the speed of the platform goes below the required minimum speed which will enable the person cross over.  while As the magnitude of the length tends towards the same magnitude of the height the speed required to cross the moat increases towards infinity and this speed cannot be attained by the person hence he will land in the moat.

Hence we can conclude that The person lands in the moat if the rope's length is very short because The speed of the platform is less than the required minimum speed  and  The person lands in the moat if the rope length is similar to the height of the platform because,the speed required to cross the moat approaches infinity.

Learn more about obstacle course : https://brainly.com/question/241926

A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.

Answers

Answer:

Explanation:

Given

Initial reading on scale =40 N

So, we can conclude that weight of the sack is 40 N

After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)

This net downward force is the resultant of earth graviational pull and the applied upward force.

So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.

This situation can be diagramatically represented by figure given below  

Answer:

40N

Explanation:

trust

Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s

Answers

Answer:

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

Explanation:

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

[tex]v = r\cdot \omega[/tex] (Eq. 1)

Where:

[tex]r[/tex] - Radius of rotation of the particle, measured in meters.

[tex]\omega[/tex] - Angular velocity, measured in radians per second.

[tex]v[/tex] - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

[tex]\omega = \frac{\theta}{t}[/tex] (Eq. 2)

Where:

[tex]\theta[/tex] - Angular displacement, measured in radians.

[tex]t[/tex] - Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

[tex]v = \frac{r\cdot \theta}{t}[/tex] (Eq. 3)

From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:

[tex]s = r\cdot \theta[/tex]

[tex]v = \frac{s}{t}[/tex]

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

plzzz helppp
You are pushing a box North in the hallway, at 20n, and a friend gets in front of the box and goes in the opposite direction, at 30n. What direction is the box going at? How much force does the box have going in that direction?​

Answers

Answer:

the box is going south at 10n

Explanation:

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