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Triss [41]
3 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]3 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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Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of
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Answer:

2.27\cdot 10^{49}

Explanation:

The gravitational force between the proton and the electron is given by

F_G=G\frac{m_p m_e}{r^2}

where

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Substituting numbers into the equation,

F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N

The electrical force between the proton and the electron is given by

F_E=k\frac{q_p q_e}{r^2}

where

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q_p = q_e = q is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N

So, the ratio of the electrical force to the gravitational force is

\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}

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In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so
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Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

a_{c}=\omega^2\times r

Put the value into the formula

a_{c}=(18.6)^2\times0.681

a_{c}=235.5\ m/s^2

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.681\times67.7

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Using formula of total acceleration

a=\sqrt{a_{c}^2+a_{t}^2}

Put the value into the formula

a=\sqrt{(235.5)^2+(46.104)^2}

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Using formula of the direction

\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})

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\theta=\tan^{-1}(\dfrac{46.104}{235.5})

\theta=11.0^{\circ}

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

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