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Triss [41]
2 years ago
6

One of the largest barometers ever built was an oil-filled barometer constructed in Leicester, England in 1991. The oil had a he

ight of 12.2 m. Assuming a pressure of 1.013 × 105 Pa, what was the density of the oil used in the barometer?
Physics
1 answer:
ZanzabumX [31]2 years ago
5 0

Answer:

ρ = 830.32 kg/m³

Explanation:

Given that

Oil head = 12.2 m

h= 12.2 m

Pressure P = 1.013 x 10⁵ Pa

Lets take density of the liquid =ρ

The pressure due to liquid P given as

P = ρ g h

Now by putting the all values in the above equation

1.013 x 10⁵ Pa = ρ x 10 x 12.2                 ( take g =10 m/s²)

ρ = 830.32 kg/m³

Therefore the density of oil is 830.32  kg/m³

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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
2 years ago
An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
7 0
2 years ago
The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal
8090 [49]
Α=2.4 \frac{m}{s}

β=1.2 \frac{m}{s^2}

x(t)=at

y(t)=3-βt^2

Vx(t)=α

Vy(t)=-2βt

vectorV=[α;-2βt]

ax(t)=0

ay(t)=-2βt

vector a [0;-2βt]


3 0
2 years ago
Calculate the pressure, in atmospheres, exerted by each of the following:
gregori [183]

Answer:

a) 14.2 atm

b) 4.46 atm

c) 1.06 atm

Explanation:

For an ideal gas,

PV = nRT

P = pressure of the gas

V = volume occupied by the gas

n = number of moles of the gas

R = molar gas constant = 0.08206 L.atm/mol.K

T = temperature of the gas in Kelvin

a) For HF,

P =?, V = 2.5L, n = 1.35 moles, T = 320K

P = 1.35 × 0.08206 × 320/2.5

P = 14.2 atm

b) For NO₂

P =?, V = 4.75L, n = 0.86 moles, T = 300K

P = 0.86 × 0.08206 × 300/4.75

P = 4.46 atm

c) For CO₂

P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K

P = 2.15 × 0.08206 × 330/55

P = 1.06 atm

4 0
2 years ago
A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu
Illusion [34]
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
3 0
2 years ago
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