Một mol khí đang ở điều kiện tiêu chuẩn thì bị nén váo một bình 5 lít. Nhiệt độ khí trong bình là 770 C. Tính áp suất khí?

Answers

Answer 1

Đáp án:

 p=5,74atm

Giải thích các bước giải:

áp suất khí là:

pV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atmpV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atm

p=5,74atm

Giải thích các bước giải:

áp suất khí là:

pV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atmpV=nRT⇒p=nRTV=1.0,082.(77+273)5=5,74atm


Related Questions

how does the property of being polar benefit the ability of organic compounds to function in the body.

Answers

Explanation:

Organic compounds essential to human functioning include carbohydrates, lipids, proteins, and nucleotides. These compounds are said to be organic because they contain both carbon and hydrogen. ... Carbohydrate compounds provide essential body fuel.

help me with this please

Answers

Answer:

Where is the Question?

Explanation:

I am glad to h help you out.

Answer:

Sorry is there any attachments to put with this question, If please publish question properly please so that brainly answerers will know what to answer.

Có bao nhiêu đặc trưng sinh lý của âm

Answers

Answer:

Ba đặc trưng sinh lí của âm là độ cao, độ to và âm sắc.

Explanation:

Under which conditions would a nucleus be least likely to undergo radioactive decay? A. When the strong nuclear forces are less than the electrostatic forces of repulsion
B. When the strong nuclear forces are greater than the electrostatic forces of repulsion
C. When the strong nuclear forces are less than the electrostatic forces of attraction
D. When the strong nuclear forces are greater than the electrostatic forces of attraction

Answers

Answer:

b is the correct answer i believe

Explanation:

the strong nuclear force encourages particles to come together so that they can nuetralize, a process the opposite of the electrostatic force of repulsion. This lets the atom stay intact because if the repulsion was more than the nuclear forces strength the atom would lose electrons and slowly decay

Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second.

Answers

Answer:

[tex]F = 14.2 N[/tex]  

Explanation:

From the question we are told that:

Mass [tex]m=0.15kg[/tex]

Radius [tex]r=0.6[/tex]

Angular Velocity [tex]\omega=2rev/s[/tex]

                            [tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]

Generally the equation for Force applied is mathematically given by

 [tex]F =mrw2[/tex]

 [tex]F=0.15*0.6* (4*x3.14^)2[/tex]

 [tex]F = 14.2 N[/tex]    

Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of fall at the earth's surface is given by g = Gm/R2 . What is the acceleration of a satellite moving in a circular orbit round the earth of radius 2R​

Answers

Explanation:

The weight of an object on the surface of the earth is equal to the gravitational force exerted by the earth on the object.

[tex]W=F_G[/tex]

[tex]mg = G \dfrac{mM}{R^2}[/tex]

which gives us an expression for the acceleration due to gravity g as

[tex]g = G\dfrac{M}{R^2}[/tex]

At a height h = R, the radius of a satellite's orbit is 2R. Then the acceleration due to gravity [tex]g_h[/tex] at this height is

[tex]mg_h = G \dfrac{mM}{(2R)^2}= G \dfrac{mM}{4R^2}[/tex]

Simplifying this, we get

[tex]g_h= G \dfrac{M}{4R^2} = \dfrac{1}{4} \left(G \dfrac{M}{R^2} \right) = \dfrac{1}{4}g[/tex]

bowling ball weighing 710 N attached to the ceiling by a rope of length 3.87 mThe ball is pulled to one side and released it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.51 m/s

What is the acceleration of the bowling ball, in magnitude and direction, at this instant?

What is the tension in the rope at this instant?

Answers

Answer:

a = 5.256 m / s²  with the vertical direction upwards and   T = 1090.8 N

Explanation:

to find the acceleration we must use Newton's second law, in the lowest part of the motion

          T - W = m a

where the acceleration is centripetal

         a = v² / r

let's calculate

         a = 4.51² / 3.87

         a = 5.256 m / s²

As the acceleration is centripetal it is directed towards the center of the circle, which in the lower part coincides with the vertical direction upwards.

Let's find the tension of the rope with the first equation

        T = W + m a

         

        W = m g

let's calculate

         T = 710 + 710 5.256 / 9.8

         T = 1090.8 N

a brick of mass 0.8 kg is accidentally dropped from a high scaffolding. it reaches the ground with a kinetic energy of 240 J. How high is scaffolding ?(Take acceleration due to gravity g be 10 m s-¹)​

Answers

Answer:

30 m

General Formulas and Concepts:

Energy

Gravitational Potential Energy: [tex]\displaystyle U_g = mgh[/tex]

m is mass (in kg)g is gravityh is height (in m)

Kinetic Energy: [tex]\displaystyle KE = \frac{1}{2}mv^2[/tex]

m is mass (in kg)v is velocity (in m/s²)

Law of Conservation of Energy

Explanation:

Step 1: Define

Identify variables

[Given] m = 0.8 kg

[Given] g = 10 m/s²

[Given] U = 240 J

[Solve] h

Step 2: Solve for h

[LCE] Substitute in variables [Gravitational Potential Energy]:                      (0.8 kg)(10 m/s²)h = 240 JMultiply:                                                                                                             (8 kg · m/s²)h = 240 JIsolate h [Cancel out units]:                                                                             h = 30 m

High speed stroboscopic photographs show that the head of a 183 g golf club is traveling at 58.6 m/s just before it strikes a 46.6 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact.

Answers

Answer:

The speed of the golf ball just after the impact is 73.04 m/s.

Explanation:

Given that,

The mass of golf club, m₁ = 183 g = 0.183 kg

The mass of golf ball, m₂ = 46.6 g = 0.0466 kg

The initial speed of golf club, u₁ = 58.6 m/s

The initial speed of a golf ball, u₂ = 0

The final speeds of club, v₁ = 40 m/s

We need to find the speed of the golf ball just after impact. Using the conservation of momentum to find it.

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.183 (58.6)-0.183(40)}{0.0466 }\\\\=73.04\ m/s[/tex]

So, the speed of the golf ball just after the impact is 73.04 m/s.

The figure shows a graph of the angular velocity of a rotating wheel as a function of time. Although not shown in the graph, the angular velocity continues to increase at the same rate until t = 8.6 s. What is the angular displacement of the wheel from 0 to 8.6 s?

Answers

Answer:

θ = 33.54 rad

Explanation:

This is a circular motion exercise

         θ= θ₀ + w₀ t + ½ α t²

suppose that for t = 0 the body is at its initial point θ₀ = 0 and from the graph we see that the initial angular velocity w₀ = -9.0 rad / s

we look for the angular acceleration,

         α = [tex]\frac{\Delta w}{ \Delta t}[/tex]

from the graph taken two points

         α = [tex]\frac{0 - (-9.0)}{3.0 - 0}[/tex]

         α = 3 rad / s²

we substitute in the first equation

            θ = 0 -9 t + ½ 3 t²

the displacement is requested for t = 8.6 s

           θ = = -9  8.6 + 3/2  8.6²

           θ = 33.54 rad

The angular displacement of the wheel from 0 to 8.6s is 33 radians.

Given to us

t = 8.6s

Acceleration of the wheel

We know acceleration is the ratio of velocity and time, therefore, the slope of the velocity-time graph will give us acceleration, therefore,

At point t=3, ω =  0

At point t = 5, ω = 6

Acceleration = slope of the Velocity-time graph = 3 rad/sec²

Angular displacement

Using the equation,

[tex]\theta = \omega_0t+\dfrac{1}{2}\alpha t^2[/tex]

SUbstitute values,

[tex]\theta = (-9.0\times 8.6)+\dfrac{1}{2}(3\times 8.6^2)\\\theta = 33\rm\ radians[/tex]

Hence, the angular displacement of the wheel from 0 to 8.6s is 33 radians.

Learn more about Angular displacement:

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The idea of a field was needed for which reason?
A. There was no way to calculate forces.
B. Forces were exerted without physical contact.
C. There was no way to know the direction of forces.
D. Forces affect only certain things.

Answers

Answer:

i believe b

Explanation:

for example, gravity works without directly contacting objects so a field helps visualize how gravity objects without physical interaction

Answer:

Explanation:

giâir

A 25N[L] force acts on a 1kg object that is already travelling at 10m/s[R]. The force acts for t=2s. What is its final speed? *
A) 25m/s
B) 25m/s/s
C) 40m/s[R]
D) 40m/s[L]
show your full work please

Answers

Answer:

C) 40m/s [R]

Explanation:

Given the following data;

Force = 25 N [L]

Mass = 1 kg

Initial velocity = 0 m/s

Velocity = 10 m/s [R]

Time = 2 seconds

To find the final velocity;

First of all, we would determine the acceleration of the object.

Force = mass * acceleration

25 = 1 * acceleration

Acceleration = 25/1

Acceleration = 25 m/s²

Next, we would determine the final velocity by using the first equation of motion;

[tex] V = U + at[/tex]

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the formula, we have;

[tex] V = 0 + 25*2[/tex]

[tex] V = 0 + 50 [/tex]

V = 50 m/s

Final velocity = 50 - 10

Final velocity = 40 m/s [R]

show your full steps

Answers

Answer:

x = 13552.6 m

Explanation:

This is a projectile throwing exercise

Let's start by looking for the time it takes for the pump to reach the ground y = 0

         y = y₀ + v_{oy} t - ½ g t²

as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t²

         t = [tex]\sqrt{2y_o/g}[/tex]

         t = [tex]\sqrt{2 \ 10000/9.8}[/tex]

         t = 45.175 s

with this time we can find the distance it travels horizontally

         x = v₀ₓ t

         x = 300 45,175

         x = 13552.6 m

the bomb must be dropped at this distance before hitting the target

Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of charge from one sphere to the other? Answer in Joules.

Answers

Answer:

Use 2>60n to get 19>089

Explanation:

cho hệ cơ học như hình vẽ hai đầu dây buộc hai vật có khối lượng tương ứng là m1=2kg và m2>m1 lấy g=10m/s sau 1s kể từ lúc bắt đầu chuyển dộng hệ vật đi được 50 cm tính m2 và sức căng của dây

Answers

xin lỗi không có sơ đồ vui lòng cho biết sơ đồ

what was the significance of jumping a.keep the snake b.keep feet cleans c.avoid the hot water d.avoid the Bumbo stick

Answers

Answer:

D I think I’m not for sure

Explanation:

HELP PLSS I CANT FAIL!!!
Elements from Period 3 of the periodic table are highlighted. Which element
is a metalloid?
A. Sodium
B. Argon
C. Sulfur
D. Silicon

Answers

Answer:

Explanation:

Look at the color scheme. That will help you a lot.

The metals are Na Mg and Al. They are colored Blue.

The Non metals are colored yellow.

Seven of the eight entries are taken up by yellow or blue. There is only 1 element left over and that is Si. So it must the metalloid. It has properties of the both the metals and the non metals.

Answer: silicon

Explanation:

5. The disks, or pads, that exist between bones as gliding joints are made of which substance?

Answers

Answer:

cartilage

Explanation:

Answer:

Cartilage

Explanation:

Correct in AE

Two
cars, A and B , are travelling in the same
direction
although
Car B is 186 metres behind A. The
speed car B is 24.4 m/s and the speed of car A is 18.6 m/ s. How much time does it take for car B to catch up with car A?

Answers

Answer:

SA = 18.6 * t + 186    total distance traveled by A in time t

SB = 24.4 t                total distance car B must travel to catch car A

SA - SB = 0 = 18.6 t + 186 = 24.4 t

(24.4 - 18.6) t = 186

t = 186 m / 5.8 m/s = 32 s

Check 18.6 * 32 + 186 = 781 m

24.4 * 32 = 781 m

PAY ATTENTION MY QUESTION ASK FOR RADIATION!!!
You sit with friends around a campfire, roasting marshmallows. Which
transfer of thermal energy involved in this system is an example of radiation

Answers

Answer:

The answer is c

Thermal energy moves within the air from the flames to the marshmallow.

Explanation:

Hope it helps

You sit with friends around a campfire, roasting marshmallows. then the transfer of thermal energy involved in this system is an example of radiation Thermal energy moves within the air from the flames to the marshmallow. Hence option C is correct.

What is thermal Energy ?

In physics and engineering, the phrase "thermal energy" is thrown around in a lot of different situations. It can relate to a variety of distinct physical notions. Included in this are the internal energy, or enthalpy, of a body of matter and radiation; heat, which is a form of energy transfer (as is thermodynamic work); and the characteristic energy of a degree of freedom in a system described in terms of its microscopic particulate constituents (where T denotes temperature and k denotes the Boltzmann constant.

Hence option C is correct.

To know more about Energy :

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#SPJ2.

(4.56 x 10^-13)-(1.17 x 10^-13)

Answers

3.39 x 10^-13

Please mark brainliest!

While taking a psychology exam, Lori got up and shut the classroom door due to outside noise that affected her concentration. Which theory explains Lori’s behavior?

Group of answer choices

Drive-reduction theory

Expectancy theory

Instinct theory

Arousal theory

Answers

Answer:

drive reduction theory

Explanation:

I would say that because of all the cars beeping and making A Lot of cachos on the street so that will definitely affect her taking her exam

Theory explains Lori’s behavior is b) Drive-reduction theory

What is Drive-reduction theory?

It is based on the idea that  the primary motivation behind all human behavior is to reduce drives . A drive is a state of  discomfort which is triggered by a person's physiology or biological need .

so , she removed discomfort that can happen by the noise outside the classroom in order to maintain student's concentration

correct answer is b) Drive-reduction theory

learn more about  Drive-reduction theory

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A single loop of wire with an area of 0.0780 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.240 T/s .

Requried:
If the loop has a resistance of 0.700Ω. Find the current induced in the loop.

Answers

Answer:

The induced current is 26.7 mA

Explanation:

Given;

area of the loop, A = 0.078 m²

initial magnetic field, B₁ = 3.8 T

change in the magnetic field strength, dB/dt = 0.24 T/s

The induced emf is calculated as;

[tex]emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V[/tex]

The resistance of the loop = 0.7 Ω

The induced current is calculated as;

[tex]V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA[/tex]

Leakage of liquid petroleum gas can be detected from a distance . name the process ​

Answers

Answer:

ultrasonic

Explanation:

they detect the acoustic emission created when a pressured gas expands in a low pressure area through a small orifice(the leak).

A 2.80 kg mass is dropped from a
height of 4.50 m. Find its potential
energy (PE) when it is 3.00 m
above the ground.

Answers

Answer:

PE = 82.32J

Explanation:

PE = m*g*h

PE = 2.80kg*9.8m/s²*3m

PE = 82.32J

Mark Brainliest please

Answer : 41.2 J

Explanation

The energy that the ball had instantaneously prior to being dropped was Potential Energy (EP) only and gravitational potential energy is calculated as Ep = mgh = (2.80)(9.81)(4.50) = 123.606 Joules.

Because no outside energy was added and no work was done, the energy instantaneously prior to being dropped is equal to the energy when the mass is 3 m above the ground only at that point, the energy consists of two parts, gravitational Potential Energy (EP) and Kinetic Energy (EK). So, considering the point just prior to the drop as point 1 and the point 3 m above the ground as point 2, we can say:
EP1 = EP2 + EK2
so EK2 = EP1 - EP2
and EP2 = (2.80)(9.81)(3.00) = 82.404
So EK2 = 123.606 - 82.404 = 41.2 Joules

A solid aluminum sphere of radius R has moment of
inertia I about an axis through its center. What is the
moment of inertia about a central axis of a solid
aluminum sphere of radius 2R?
1. 21
2. 41
3. 87
4. 161
5. 321​

Answers

Answer:

5. 32I

Explanation:

The moment of inertia of a solid sphere about its central axis is given by

I = [tex]\frac{2}{5} MR^2[/tex]            ------------------(i)

Where;

M = mass of the sphere

R = radius of the sphere.

From the question;

Case 1: The aluminum sphere has a radius R and moment of inertia I.

This means that we can substitute these values of R and I into equation (i) and get;

I = [tex]\frac{2}{5} MR^2[/tex]       --------------(ii)

M is the mass of the aluminum sphere and is given by;

M = pV

Where;

p = density of aluminum

V = Volume of the sphere = [tex]\frac{4}{3} \pi R^3[/tex]

=> M = p([tex]\frac{4}{3} \pi R^3[/tex])              --------------------(*)

Case 2: An aluminum sphere with a radius of 2R instead.

Let the moment of inertia in this case be I' and mass be M'

Substituting R = 2R, M = M' and I = I' into equation (i) gives

I' = [tex]\frac{2}{5} M'(2R)^2[/tex]       ------------------(iii)

Where;

M' = pV'

p = density of aluminum

V' = volume of the sphere = [tex]\frac{4}{3} \pi (2R)^3[/tex]

=> M' = p([tex]\frac{4}{3} \pi (2R)^3[/tex])

Rewriting gives;

M' = p([tex]\frac{4}{3} \pi (2)^3(R)^3[/tex])

M' = p([tex]\frac{4}{3} \pi8(R)^3[/tex])

M' = 8p([tex]\frac{4}{3} \pi R^3[/tex])

From equation (*), this can be written as

M' = 8M

Now substitute all necessary values into equation (ii)

I' =  [tex]\frac{2}{5} M'(2R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(2R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(2)^2(R)^2[/tex]  

I' =  [tex]\frac{2}{5} (8M)(4)(R)^2[/tex]

I' =  [tex]\frac{2}{5} (32M)(R)^2[/tex]

I' =  [tex]32[\frac{2}{5}MR^2][/tex]

Comparing with equation (ii)

I' =  [tex]32[I][/tex]

Therefore, the moment of inertia about a central axis of a solid

aluminum sphere of radius 2R is 32I

f(x)=
[tex]f{x} = \sqrt{x} [/tex]

Answers

What’s the question about ?

you are traveling in a convertible with the top down. the car is moving at a constant velocty of 25 m/s due east along falt ground. you throw a tennis ball straight upward at a speed of 15 m/swhen the ball just leaves your hand, what is the speed of the ball as measured by an observer on the side of the road

Answers

Answer:

Teh velocity of ball with respect to the ground is 29.2 m/s.

Explanation:

Velocity of car, vc = 25 due m/s east

velocity of ball with respect to car, v(b,c) = 15 m/s due Z axis

Write the velocities in the vector form

[tex]\overrightarrow{v}_{c}=25 \widehat{i}\\\\\overrightarrow{v}_{(b,c)}=15 \widehat{k}[/tex]

The velocity of ball with respect to ground is

[tex]\overrightarrow{v}_{(b,c)}= \overrightarrow{v}_{(b,g)}-\overrightarrow{v}_{(c,g)}\\\\15 \widehat{k} = \overrightarrow{v}_{(b,g)} - 25 \widehat{i}\\\\\overrightarrow{v}_{(b,g)} = 25 \widehat{i} + 15 \widehat{k}\\\\v_{(b,g)}=\sqrt{25^2 + 15^2}=29.2 m/s[/tex]

A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 70.0 Hz, amplitude 5.00 mm, and wavelength 0.600 m. (a) How long does it take the wave to travel a distance of 8.00 m along the length of the string

Answers

Answer:

The time taken by the wave to travel 8 m is 0.19 s .

Explanation:

frequency, f = 70 Hz

Amplitude, A = 5 mm

Wavelength = 0.6 m

The wave speed is given by

wave speed, v = frequency x wavelength

v = 70 x 0.6

v = 42 m/s

Distance, d = 8 m

Let the time taken is t.

So, [tex]t =\frac{d}{v}\\\\t =\frac{8}{42}\\\\t = 0.19 s[/tex]

   

When a Frisbee is flying horizontally through the air, the speed of the air flowing over the Frisbee's top is

Answers

Group of answer choices.

A. higher than the speed of the air beneath it and the pressure on top of the Frisbee is greater than the pressure beneath it.

B. lower than the speed of the air beneath it and the pressure on top of the Frisbee is less than the pressure beneath it.

C. lower than the speed of the air beneath it and the pressure on top of the Frisbee is greater than the pressure beneath it.

D. higher than the speed of the air beneath it and the pressure on top of the Frisbee is less than the pressure beneath it.

Answer:

D. higher than the speed of the air beneath it and the pressure on top of the Frisbee is less than the pressure beneath it.

Explanation:

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

[tex] Acceleration = \frac {Net \; force}{mass} [/tex]

When a frisbee flies horizontally through the air, it experiences air above and beneath it. Also, there is a pressure acting on both on top and below the frisbee.

Typically, the air flowing over the top of the frisbee has a speed that is higher than the speed of the air beneath it.

Furthermore, the frisbee is able to remain in flight because the pressure on top of the frisbee is less than the pressure beneath it.

Other Questions
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