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nexus9112 [7]
3 years ago
11

A closed container is filled with oxygen. the pressure in the container is 405 kpa . what is the pressure in millimeters of merc

ury? express the pressure numerically in millimeters of mercury.
Physics
1 answer:
kari74 [83]3 years ago
8 0

1 mm Hg pressure is the pressure due to 1 mm height of mercury

so it is given as

P = \rho g h

P = 1 * 10^{-3} * 13700 * 9.8 = 134.26 Pa

now we know that pressure is

P = 405 kPa

P = 405 * 10^3 Pa

so in mm Hg this pressure is

P = \frac{405 * 10^3 }{134.26} = 3016.5 mm Hg

so pressure in millimeter Hg will be 3016.5 mm Hg

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A thin film of cooking oil ( n = 1.47 ) is spread on a puddle of water ( n = 1.35 ) . What is the minimum thickness D min of the
zlopas [31]

Answer:

The minimum thickness of the oil is 77.55 nm

Explanation:

Given:

Refractive index of oil n_{o} = 1.47

Refractive index of water n_{w} = 1.35

Wavelength of light \lambda= 456 \times 10^{-9} m

From the equation of thin film interference,

The minimum thickness is given by,

    2n_{o} t = (n+\frac{1}{2}) \lambda

Where n = 0,1,2,3.........,t = thickness

Here we have to find minimum thickness so we use n = 0

     2n_{o} t  =( 0+\frac{1}{2} )\lambda

   t = \frac{\lambda }{4 n_{o} }

   t = \frac{456 \times 10^{-9} }{4 \times 1.47}

   t = 77.55 \times 10^{-9} m

   t  = 77.55 nm

Therefore, the minimum thickness of the oil is 77.55 nm

7 0
4 years ago
German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity (
Ierofanga [76]

Answer:

The uncertainty in the position of the electron is 5.79x10^{-9}m

Explanation:

The Heisenberg uncertainty principle is defined as:

\Lambda p\Lambda x ≥ \frac{h}{4 \pi}  (1)

Where \Lambda p is the uncertainty in momentum, \Lambda x is the uncertainty in position and h is the Planck's constant.

The momentum is defined as:

p =mv  (2)

Therefore, equation 2 can be replaced in equation 1

\Lambda (mv) \Lambda x ≥ \frac{h}{4 \pi}

Since, the mass of the electron is constant, v will be the one with an associated uncertainty.

m \Lambda v \Lambda x ≥ \frac{h}{4 \pi} (3)

Then, \Lambda x can be isolated from equation 3

\Lambda x ≥ \frac{h}{m \Lambda v 4 \pi}  (4)

\Lambda x = \frac{6.626x10^{-34}J.s}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) 4 \pi}

But 1J = Kg.m^{2}/s^{2}

\Lambda x = \frac{(6.624x10^{-34} Kg.m^{2}/s^{2}.s)}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) (4 \pi)}

\Lambda x = 5.79x10^{-9}m

Hence, the uncertainty in the position of the electron is  5.79x10^{-9}m

7 0
3 years ago
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marta [7]

Answer:

A

My logic:

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8 0
3 years ago
A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.
miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

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\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

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Everyday activities do NOT produce significant muscle growth because:
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Answer:

When elements bond together or when bonds of compounds are broken and form a new substance

Explanation:

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