We do not use water instead of mercury in a barometer​

Answers

Answer 1

Answer:

Because

1)Water is relatively less dense. Mercury is 13.6 times more dense than water. SO, If you use water, you have to have the length of barometer of length (or height) 13.534 times the length of mercury barometer, which may be more than 11 meter in length.

2)Also mercury as compared to water, has comparatively less specific heat and good conductor of heat, could come to the same temperature of the atmosphere more quickly.

Answer 2

Answer:

The atmospheric pressure at sea level 76 cm of Hg=1.013× 10⁵ pascal .

Explanation:

That is if water is used in barometer tube instead of mercury the length of the tube must be greater than 10.326 cm.so we cannot replace mercury by water in the barometer.


Related Questions

A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz

Answers

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

Given the following data:

Period = 645 μs

Note: μs represents microseconds.

Conversion:

1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds

645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

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

Substituting the value into the formula, we have;

[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

Conversion:

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = [tex]0.001[/tex] × [tex]1550.39[/tex]

X = 155039 kHz

To 3 significant figures;

Frequency = 155 kHz

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

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Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]​

Answers

Answer:

Gap left = Change in length on heating

Gap=Initial length×Coefficient of linear expansion×change in temperature

Gap=10×0.000012×15m

⟹Gap=0.0018 m

this is an example u have to put your equation in it

A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?

Answers

Answer:

0

Explanation:

The speed of the ball when it reaches the floor is 0 because any object at rest or in uniform motion has no speed or velocity

Which of the following accurately describes circuits?
options:

A)

In a parallel circuit, there's only one path for the current to travel.

B)

In a series circuit, the amount of current passing through each part of the circuit may vary.

C)

In a series circuit, the current can flow through only one path from start to finish.

D)

In a parallel circuit, the same amount of current flows through each part of the circuit.

Answers

' C ' is the only correct statement.

3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic

Answers

The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

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Cho dòng điện xoay chiều trong sản xuất và sinh hoạt ở nước ta có tần số f = 50Hz. Tính chu kỳ T và tần số góc ω?

Answers

Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

what Is accuracy ............​

Answers

Answer:

Accuracy is how much the consequence of an estimation adjusts to the right worth or a norm' and basically alludes to how close an estimation is to its concurred esteem

《OAmalaOHopeO》

Answer:

In a set of measurements, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.

Explanation:

_Hope it helps you_

Question 5 of 10
What must be the same for two resistors that are connected in parallel?

Answers

Answer:

in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .

Explanation:

Write the derivation and unit of impulse​

Answers

Answer:

impulse applied to an object produces an equivalent vector change in its linear momentum, also in the resultant direction. The SI unit of impulse is the newton second (N⋅s), and the dimensionally equivalent unit of momentum is the kilogram meter per second (kg⋅m/s).

The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m

Answers

Answer:

[tex]v=(6ti+6k)\ m/s[/tex]

Explanation:

Given that,

The position of a particle is given by :

[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]

Let us assume we need to find its velocity.

We know that,

[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]

So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].

A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.

Answers

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}[/tex]

Use the formula of magnification

[tex]m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft[/tex]

A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m>s. What is the maximum temperature increase that the water could have as a result of this event

Answers

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

why is nut-cracker 2nd class lever?​

Answers

2nd class leaver refers to such leaver in which load lies between effort and fulcrum.In a nut cracker too load is in between effort and fulcrum.Thus, nut cracker is a 2nd class leaver.......

At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder arrive 7.5 seconds later. How many miles away was the lighting strike? (Assume the light takes essentially no time to reach you).

Answers

Answer:

1.6031 miles

Explanation:

Given the following data;

Speed = 344 m/s

Time = 7.5 seconds

To find how many miles away was the lighting strike;

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 344 * 7.5

Distance = 2580 meters

Next, we would have to convert the value of the distance travelled in meters to miles;

Conversion:

1609.344 metres = 1 mile

2580 meters = X mile

Cross-multiplying, we have;

X * 1609.344 = 2580

X = 2580/1609.344

X = 1.6031 miles

15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?

Answers

Answer: (15 - 0)/1.8 = 8. 33m/s^2

Explanation:

The acceleration of the racehorse is 8.33 m/s²

The given parameters;

initial velocity of the racehorse, u = 0

final velocity of the racehorse, v = 15  m/s

time of motion of the horse, t = 1.8 s

The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;

[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]

Thus, the acceleration of the racehorse is 8.33 m/s²

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A 2.5 kg block slides along a frictionless surface at 1.5 m/s.A second block, sliding at a faster 4.1 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/s. What was the mass of the second block?

Answers

Answer:

1.5kg

Explanation:

Given data

mass m1= 2.5kg

mass m2=??

velocity of mass one v1= 1.5m/s

velocity of mass two v2= 4.1m/s

common velocity after impact v= 2.5m/s

Let us apply the formula for the conservation of linear momentum for inelastic collision

The expression is given as

m1v1+ m2v2= v(m1+m2)

substitute

2.5*1.5+ m2*4.1= 2.5(2.5+m2)

3.75+4.1m2= 6.25+2.5m2

collect like terms

3.75-6.25= 2.5m2-4.1m2

-2.5= -1.6m2

divide both sides by -1.6

m2= -2.5/-1.6

m2= 1.5 kg

Hence the second mass is 1.5kg

Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for longer wavelengths of light. The work function of the surface is

Answers

Explanation:

Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]

[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]

[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]

The work function [tex]\phi[/tex] is then

[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]

[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]

The work function of the surface is equal to 3.98 × 10⁻¹⁹J.

What are frequency and wavelength?

The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.

The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.

The expression between wavelength (λ), frequency, and speed of light (c) is:

c = νλ

Given, the wavelength of the light, ν = 500 nm

The frequency of the light can determine from the above-mentioned relationship:

ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz

The work function = h ν =  6 × 10¹⁴ × 6.626 × 10⁻³⁴

φ = 3.98 × 10⁻¹⁹J

Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.

Learn more about wavelength and frequency, here:

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Flapping flight is very energy intensive. A wind tunnel test
on an 89 g starling showed that the bird used 12 W of
metabolic power to fly at 11 m/s. What is its metabolic power for starting flight?

Answers

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=[tex]\frac{P}{m}[/tex]

Using the formula

Metabolic power for starting flight=[tex]\frac{12}{0.089}[/tex]

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

What is the momentum of a 36.9 N bowling ball with a velocity of 7.56 m/s?

Answers

Answer:

momentum (m)=36.9N

velocity (v)=7.56m/s

now,

momentum (m)=m×v

36.9=m×7.56

36.9÷7.56=m

m=4.89kg

A rock is suspended by a light string. When the rock is in air, the tension in the string is 37.8 N. When the rock is totally immersed in water, the tension is 32.0 N. When the rock is totally immersed in an unknown liquid, the tension is 20.2 N. What is the Density of the unknown liquid?

Answers

When the rock is suspended in the air, the net force on it is

F₁ = T₁ - m₁g = 0

where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So

T₁ = m₁g = 37.8 N

When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then

F₂ = T₂ + B₂ - m₁g = 0

where B₂ is the magnitude of the buoyant force. Then

B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N

B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then

5.8 N = m₂g   ==>   m₂ ≈ 0.592 kg

If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was

1.00 × 10³ kg/m³ = m₂/V   ==>   V ≈ 0.000592 m³ = 592 cm³

and this is also the volume of the rock.

When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is

F₃ = T₃ + B₃ - m₁g = 0

which means the buoyant force is

B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N

The mass m₃ of the liquid displaced is then

17.6 N = m₃g   ==>   m₃ ≈ 1.80 kg

Then the density ρ of the unknown liquid is

ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³

Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed​

Answers

Answer:

c.

magnitude of displacement

Two infinitely long parallel wires carry current in opposite directions. Wire 1 has current 15.0 A and wire 2 has current 19.9 A. If they are separated by a distance 4.3 m, at what location between the wires is the net magnetic field be twice the strength of the magnetic field from wire 1

Answers

Answer:

x= 2*I1*d/(I1+I2) meter

What is the need for satellite communication elaborate

Answers

The high frequency radio waves used for telecommunications links travel by line of sight and so are obstructed by the curve of the Earth. The purpose of communications satellites is to relay the signal around the curve of the Earth allowing communication between widely separated geographical points.

Explanation:

hope it helps!!

1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s , después cambia su trayectoria cuando es lanzada hacia arriba 25m durante 7 s. Calcular la velocidad y la rapidez al punto final (altura maxima) al que llegó la pelota.

2. Una mariposa vuela en línea recta hacia el sur recorriendo una distancia de 15 m durante 28 s, después cambia de dirección hacia el Oeste recorriendo una distancia de 50 m en un tempo de 80 s ¿cuál es la velocidad y rapidez de la mariposa?

3.- Una persona camina durante 21 minutos hacia el este de su casa una distancia de 1500 m y después cambia su dirección hacia el Norte recorriendo una distancia de 3350 m en un tiempo 32 minutos llegando al supermercado. ¿Calcula la velocidad y rapidez de la persona?

4.- Un automóvil se mueve al Oeste recorriendo una distancia de 80 km en 1.2 horas, posteriormente cambia su trayectoria hacia el Sur, recorriendo una distancia de 120 km en un tiempo 1.6 hora. ¿Calcula la velocidad y rapidez del automóvil?

Answers

Answer:

https://youtu.be/ymHHdoCGJOU

(a+b)[tex]x^{2}[/tex]

Answers

u distribute it
(a+b) • x2
ax2 + bx2
THE ANSWER IS ax2 + bx2 (the two means squared)

Solve numerical problem. Please give me step - step explanation Help me out plz

Answers

Answer:

You should multiply 60 kg*9.8 and answer will come.

Hope this will help you.

Answer:

yes she is right you should multiple 60*9.8

have a great day God bless you

What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm

Answers

Hi! The answer is 15km I'm pretty sure

Answer:

The submarine's maximum safe depth in sea water is 801.678 m.

Explanation:

P=Po+(rho)*g*h

Max Pressure = Initial Pressure + (Water Density)(Gravity)(Max Depth)

Area of Window = Pi*(Diameter/2)^2 = Pi*(.4m/2)^2 = 0.125664 m^2

Max Pressure= (1.0*10^6 N)/(0.125664 m^2)= 7.95775-E6 Pa

Initial Pressure= 1atm= 101.3kPa= 101300Pa

Water Density (rho) = 1000kg/m^3

Gravity= 9.8m/s^2

So rearranging for h= (P-Po)/((rho)*g)

h=((7.95775-E6Pa)-(101300Pa))/((1000kg/m^3)(9.8m/s^2))= 801.678 m

What does E=mc2 stand for?

Answers

It stands for energy=mass times the speed of light squared.

That mean mass-energy equivalence

A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.

Answers

Answer:

the distance from the cornea where a very distant object will be imaged is 23.35 mm

Explanation:

Given the data in the question;

For a spherical refracting surface;

[tex]n_i[/tex]/[tex]d_0[/tex] + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

where [tex]n_i[/tex] is the index of refraction of the light of ray in the incident medium

[tex]d_0[/tex] is the object distance

[tex]n_t[/tex] is the index of refraction of light ray in the refracted medium

[tex]d_i[/tex] is the image distance

R is the radius of curvature

Now, let [tex]d_0[/tex] = ∞, such that;

[tex]n_i[/tex]/∞ + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

0 + [tex]n_t[/tex]/[tex]d_i[/tex] = ( [tex]n_t[/tex] - [tex]n_i[/tex] )/R

we make [tex]d_i[/tex] subject of the formula

[tex]n_t[/tex]R = [tex]d_i[/tex]( [tex]n_t[/tex] - [tex]n_i[/tex] )

[tex]d_i[/tex] = ( [tex]n_t[/tex] × R ) / ( [tex]n_t[/tex] - [tex]n_i[/tex] )

given that; R = 6.50 mm, [tex]n_t[/tex] = 1.41, we know that [tex]n_i[/tex] = 1.00

so we substitute

[tex]d_i[/tex] = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )

[tex]d_i[/tex] = 9.165 / 0.41

[tex]d_i[/tex] = 23.35 mm

Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm

A) In terms of electrolysis, it’s been said from multiple sources online that “Using water's density and relative atomic populations, it is estimated by a mass balance that approximately 2.38 gallons of water are consumed as a feedstock to produce 1 kg of hydrogen gas (14.13 liters), assuming no losses.”

B) However, 1 Gallon of water is said to contain approximately 4,707 liters of hydrogen.

How can both statements be correct under normal atmospheric conditions, since even with 80% efficiency of current PEM electrolyzers the first statement (A) is nowhere near the +4,000 liters of the second approximation (B)?

Answers

Answer:

hhhhhhhjjjkkllkcftkbgfjknhglncg

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