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muminat
3 years ago
5

He instruction booklet for your pressure cooker indicates that its highest setting is 11.1 psi . you know that standard atmosphe

ric pressure is 14.7 psi, so the booklet must mean 11.1 psi above atmospheric pressure. at what temperature in degrees celsius will your food cook in this pressure cooker set on "high"?
Physics
1 answer:
blagie [28]3 years ago
5 0

Given that,

Atmospheric Pressure = 14.7 psi

Cooking Pressure = 14.7 +11.1 = 25.8 psi

Take, Atmospheric Temperature = 25 °C

Cooking Temperature = ??

Since, we know that Gas equation is given by:

PV = nRT

or

P ∝ T

P1 / T1 = P2 / T2

14.7/ 25 = 25.8/ T2

T2 = 25*25.8/14.7

T2 = 43.87 °C

The cooking pressure will be 43.87 °C.

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Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

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\cot\phi = (1+\frac{hf}{m_{e}c^{2}  })\tan\frac{\theta }{2}      ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, m_{e} is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ      ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of  λ .

\cot\phi = (1+\frac{h}{m_{e}c\lambda  })\tan\frac{\theta }{2}

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for m_{e}, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ  and 134° for θ in the above equation.

\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9}  })\tan\frac{134 }{2}

\cot\phi=2.37

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3 years ago
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8 0
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2 years ago
The input work done on a machine is 9.63 × 103 joules, and the output work is 3.0 × 103 joules. What is the percentage efficienc
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Answer:  31%
5 0
3 years ago
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