The pressure at the ice point for a constant-volume gas thermometer is 4.81 x 10^4Pa.
While that at the steam point is 6.48 x 10^4 Pa.
What pressure would the thermometer indicate at 50°c?

Answers

Answer 1

Answer:

0 deg C = 4.81E4     pressure at 0 deg

100 deg C = 6.48E4     pressure at steam point

100 deg C - 50 deg C = (6.48 - 4.81) * 10^4 = 1.67E4 Pa

50 deg C = 50 / 100 * 1.67E4 + 4.81E4 = 5.65E4 Pa Just the halfway point between the two given pressures


Related Questions

please answer
during a journey, a car travels at 40 km in 2.5 hours, next 62 km in 3 hours, then took a break for 30 minutes, again travelled the last 120 km in 3.2 hours. calculate the average speed of the car during the journey.

Answers

average speed of the car is 23.9 km/h

Angie walked a distance of 90 meters east in 70 seconds. What was her
velocity?
A. 0.78 m/s east
B. 1.3 m/s east
O C. 7 m/s east
D. 9 m/s east

Answers

Answer:

1.3 m/s east

Step-by-step explanation:

90/70 = 1.3

Her velocity is 1.3

difference between effort distance and load distance​

Answers

Answer:

Lever systems are simple machines that change or increase the input force that we apply to the load. The lever provides us with some mechanical...

Answer:

● Effort arm or Effort distance (ED): The perpendicular distance from the fulcrum to the point of effort is called effort arm.

● Load arm or Load distance (LD): The perpendicular distance from the fulcrum to the point of load is called load arm.

think about the objects which acts as concave or convex Mirrors in your surroundings write a report on it​

Answers

Answer:

Explanation:

S.No Concave                      Mirror  Convex Mirror  

1.  Dentist's mirror               Rear view mirror of vehicles  

2.  Shaving Mirror                Streetlights  

3.  Headlight of vehicles      Water glass surface  

4.   Caving side of spoon     Curving side of spoon

5.   Torch glass                    Electric Bulb  

6.  Inner surface of              Outer surface of steel basin  

steel basin  

7.  Lunch plates                   Calling bell

8.  Microscope                       Optical instruments  

9.   Satellite dishes              Fish eye lenses

Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iron

Answers

Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up

State the relative position for the earth and sun in a lunar eclipse (in a partial and total eclipse)

Answers

Answer:

A lunar eclipse is when the Earth passes between the moon and the sun, casting a shadow on the moon. This can only occur when the sun, Earth and moon are aligned exactly, or very closely so, with the Earth in the middle.

Can someone do this for the football
Estimated volume
(cm3)
Estimated density
(g/cm3)
edge.

Answers

G/292 cm
is the answer

Answer:

0.10 g/cm3

TRUST ME GUYS

Có một số điện trở giống nhau R0 = 3

. Cần ít nhất bao nhiêu điện trở để có một
đoạn mạch có điện trở Rtđ = 8

Answers

Answer:

hlo

Explanation:

hlo olz mark me as brainlest

write down any 5 example of conservation of momentum?​

Answers

Answer:

1) Motion of air mass moving from equator northward (closer to earth axis)

2) Motion of object in orbit

3) Collision of 2 objects

4) Skater changing rotation by extension of arms

5) Motion of rocket due to velocity of expelled gas

Một vật không mang điện sẽ bị nhiễm điện dương khí

Answers

Answer:

không có điện

Explanation:

An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)

transition metals
noble gases
metalloids
alkali metals

Answers

Answer:

Alkali metals

Explanation:

Elements in this group are highly reactive, soft, lustrous and highly conductive.

An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.

Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.

This causes them to be soft and highly reactive because:

The single valance electron leads to weak bonds amongst the element's atoms which makes them softThe elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.

Examples of alkali electrons include:

Lithium Sodium Potassium etc

In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.

Find out more at https://brainly.com/question/18722874.

4. When setting goals, you should do everything EXCEPT which of the following?
Take into account your current level of activity.
Ask all of your friends what they think your goals should be..
Set a time limit.
Be realistic.

Answers

Ask all of your friends what your goals should be. These are your goals, and they are for you only. They shouldn’t be what other people think they should be, they should be what you want them to be.

Question in the picture please help me...
Show all steps please....​

Answers

Answer: i)A to B : (ice) freezing

ii) B to C (water) boiling

C to D  (steam) evaporating

explanation:  0° is the freezing point of water  when temperature increases from 0° the water starts melting. As 100° is the boiling point of water so at 100° the water  completely melts and it starts boiling during boiling water changes into steam(water vapour) and it evaporates

What is your wheel and axle

Answers

Explanation:

The wheel and axle is a type of simple machine used to make tasks easier in terms of manipulating force by applying the concept of mechanical advantage.

Động vật nào sau đây máu đi nuôi cơ thể không pha trộn giữa máu giàu O2 và máu giàu CO2?

A.
Bò sát, chim, thú

B.
Cá, bò sát, chim

C.
Cá, lưỡng cư, bò sát

D.
Cá, chim, thú

Answers

Answer:

D

Explanation:

Cá xương, chim, thú, cá sấu không có sự pha trộn máu giàu O2 và máu giàu CO2 ở tim vì tim cá có 2 ngăn, tim các loài chim, thú, cá sấu có 4 ngăn

Which element would have the lowest electronegativity? (1 point)

an element with a small number of valence electrons and a large atomic radius

an element with a small number of valence electrons and a small atomic radius

an element with a large number of valence electrons and a small atomic radius

an element with a large number of valence electrons and a large atomic radius

Answers

Answer:

an element with a small number of valence electrons and a large atomic radius

An oscillator completes 240 cycles in 5.2 minutes.
Calculate its period (in seconds) and frequency (in Hz).

Answers

Answer:

I. Period = 1.3 seconds

II. Frequency = 0.769 Hertz

Explanation:

Given the following data;

Number of oscillation = 240 cycles

Time = 5.2 minutes.

Conversion:

1 minute = 60 seconds

5.2 minutes = X seconds

X = 60 * 5.2

X = 312 seconds

To find the following;

I. Period

Mathematically, the number of oscillation of a pendulum is given by the formula;

[tex] Number \; of \; oscillation = \frac {Time}{Period} [/tex]

Making period the subject of formula, we have;

[tex] Period = \frac {Time}{Number \; of \; oscillation} [/tex]

Substituting into the formula, we have;

[tex] Period = \frac {312}{240} [/tex]

Period = 1.3 seconds

II. Frequency

[tex] Frequency = \frac {1}{Period} [/tex]

Substituting the values into the formula, we have;

[tex] Frequency = \frac {1}{1.3} [/tex]

Frequency = 0.769 Hertz

If 20N force produces an acceleration of 5ms^-2 In a body then the mass of the body will be:
A.4kg
B.5kg
C.1/4kg
D.1/5kg

Answers

F = ma
m = F/a
m = 20N / 5m/s^2
m = 4kg

Can someone pls help, thank you in advance!
What is an example of a force applied at an angle to displacement

Answers

Answer:

  an object sliding down hill

Explanation:

On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.

Two football players run towards each other along a straight path in Penrith Park in the clash between the Melbourne storms and the Penrith Panthers a month ago. Melbourne's Justin Olam who is about 95kg and ran towards Viliame Kikau at 3.75m/s. Viliame Kikau is 111kg and moves towards Justin Olam at 4.10m/s. They end up in a head-on collision and are stuck together.

A) What is their velocity immediately after the collision?

B) What are the initial and final kinetic energies of the system?​

Answers

Answer:

a) v = 0.4799 m / s,  b)  K₀ = 1600.92 J,    K_f = 5.46 J

Explanation:

a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.

Initial instant. Before the crash

        p₀ = m v₁ + M v₂

where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive

Final moment. After the crash

       p_f = (m + M) v

as the system is isolated, the moment is preserved

       p₀ = p_f

       m v₁ + M v₂ = (m + M) v

       v =[tex]\frac{m v_1 + M v_2}{m+M}[/tex]

let's calculate

        v = [tex]\frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}[/tex]

        v = 0.4799 m / s

b) let's find the initial kinetic energy of the system

         K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2

         K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2

         K₀ = 1600.92 J

the final kinetic energy

         K_f = ½ (m + M) v ^ 2

         k_f = ½ (95 + 111) 0.4799 ^ 2

         K_f = 5.46 J

1) The position of an object to the north of a flagpole is given by x(t) = bt2 – c , where b and c are constants.
a) What is v(t), the velocity of the object as a function of time?
b) What is a(t), the acceleration of the object as a function of time?
c) At some time t the object is located at the flagpole. What is the velocity of the
object at that instant?

Answers

Answer:

a) The velocity of the object as a function of time, v(t) is 2·b·t

b) The acceleration of the function of time, a(t) is 2·b

c) The time at which the object is at the flagpole is t = √(c/b)

Explanation:

The function that gives the position of the object north of the flagpole, x(t) is presented as follows;

x(t) = b·t² - c (b and c are constants)

a) The velocity of the object as a function of time, v(t), is derived as follows

v(t) = x'(t) = d(b·t² - c)/dt = 2·b·t

The velocity of the object as a function of time, v(t) = 2·b·t

b) The acceleration of the function of time, a(t) = v'(t) = d(2·b·t)/dt = 2·b

c) The time at which the object is at the flagpole is given by the x-intercept of the function, where x(t) = 0, as follows;

At the x-intercept, we have, x(t) = 0 and x(t) = b·t² - c

∴ 0 = b·t² - c, which gives

b·t² = c

t² = c/b

t = ±√(c/b), we reject the negative value to get;

The time at which the object is at the flagpole, t = √(c/b).

A cyclist goes round a circular path of circumference 343 m in s. The angle made by him, with the vertical is
WITH STEPS PLZ

Answers

you need the number of seconds to calculate the distance. I'm assuming that the seconds were written in the question but you forgot to write the here, you can calculate the distance then and find the exact point where the cyclist stopped since you have the circumference. then you can find the angle (after finding the distance he cycled) by subtracting it from 360 since the circular path will have an angle of 360 degrees.

Although your question lacks some data A general answer is provided :

The angle made vertically  = 360° - x°

where x = angle at which the the cyclist stopped

and circumference of the circular path = Total Distance travelled by the cyclist

First step : determine the distance travelled by the cyclist vertically

circumference = 2*π*r

343 = 2*π * r

∴ r = 343 / ( 2π ) = 54.59

therefore distance travelled by the cyclist vertically = 54.59 * 2 = 109.18 m

Given that the time travelled is missing

assuming the angle to the distance travelled by the cyclist before it stopped vertically = 360° - x°  

This is because the Total angle of a circular path = 360°

Learn more about circular paths : https://brainly.com/question/24210487

An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?

Answers

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

0=30^2+2*(-9.8)*x

x=-30^2/2*-9.8

x=45.92m

Answer:

35

Explanation:

dfddffffffffffffdddfr

what is Newton's first law of motion?
EXPLAIN WITH SOME EXAMPLES​

Answers

Answer:

The tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia.

Example.: A ball at rest on the ground continues to be at rest unless someone kicks it or any external force acts on it.

Explanation:

I hope this will help you buddy

7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.

Answers

Answer:

8392

Explanation:

d=s/t

The velocity ratio of a pulley system is 4. What does it means

Answers

Answer:

Velocity ratio = distance travelled by effort / distance travelled by load. Which is equal to effort arm / load arm. When velocity ratio (VR) is 4, the effort arm is greater than load arm.

A object of mass 200kg is pushed from rest by a force of 500N along a horizontal plane for 5.0 seconds. Calculate the acceleration of the object

Answers

Answer:

force=mass×acceleration

hence

acceleration is given by force÷mass

(500÷200)*5=12.5

a box being pushed with a force of 85N right slides along the floor with a constant speed of 0.5m/s. What is the force of sliding friction on the box?

Answers

Answer:

[tex]85\; \rm N[/tex].

Explanation:

The box is sliding with a constant speed in a fixed direction (to the right.) In other words the velocity of this box is constant. Hence, this box would be in a translational equilibrium. The acceleration of this box would be zero.

By Newton's Second Law of motion, the net force on this box would be [tex]0[/tex]. In other words, forces on this box are balanced.

The question is asking for the size of the friction on the box. Assuming that the floor is horizontal. The friction on this box would also be horizontal,

The only other force that could balance that friction would be the [tex]85\; \rm N[/tex] push to the right. The direction of this push is horizontal (to the right.) Hence, the entirety of that [tex]85\; \rm N\![/tex] would be in the horizontal direction.

Thus, forces on this box in the horizontal direction would be:

The [tex]85\; \rm N[/tex] push to the right.Friction that opposes the rightward motion of the box (that is, to the left.)

Since these two forces must balance each other, the size of the friction would also be [tex]85\; \rm N[/tex].

The Weight of an object is more at pole and less at equator of the earth. Give reason​

Answers

The Weight of an object is more at pole and less at equator of the earth because the distance of the pole is less than the distance of the equator from the centre of the earth.

an electron is moving at 3.6 x 10^3 m/s. a photon with what wavelength would have the same momentum????

Answers

Answer:

The wavelength of the photon that would have the same momentum as the electron is 202.2180996 nm

Explanation:

The velocity of the electron, v = 3.6 × 10³ m/s

The momentum of an electron, [tex]p_e[/tex] = m × v

Where;

v = The mass of the electron = 9.109 × 10⁻³¹ kg

∴ [tex]p_e[/tex] = 9.109 × 10⁻³¹ kg × 3.6 × 10³ m/s = 3.27924 × 10⁻²⁷ kg·m/s

According to the de Broglie equation, the momentum of a photon, p, is given as follows;

p = h/λ

Where;

h = 6.63 × 10⁻³⁴ J·s

λ = The wavelength of the photon

∴ λ = h/p

According to the question, we have;

p = [tex]p_e[/tex] = 3.27924 × 10⁻²⁷ kg·m/s

∴ λ = 6.63 × 10⁻³⁴ J·s/(3.27924 × 10⁻²⁷ kg·m/s) = 2.02180993 × 10⁻⁷ m

The wavelength of the photon, λ = 2.02180993 × 10⁻⁷ m = 202.2180993 × 10⁻⁹ m = 202.2180993 nm.

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