Apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 180kPa, T = -8.0°C = 265.15K
Final P and T values:
P = 245kPa, T = ?
Set the initial and final P/T values equal to each other and solve for the final T:
180/265.15 = 245/T
T = 361K
Answer:
Hope the above picture might help you :)
Answer:
![125\sqrt[4]{8}](https://tex.z-dn.net/?f=125%5Csqrt%5B4%5D%7B8%7D)
Explanation:
A number of the form

can be re-written in the radical form as follows:
![\sqrt[n]{a^m}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D)
In this problem, we have:
a = 1,250
m = 3
n = 4
So, if we apply the formula, we get
![1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}](https://tex.z-dn.net/?f=1%2C250%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%3D%5Csqrt%5B4%5D%7B%281%2C250%29%5E3%7D)
Then, we can rewrite 1250 as

So we can rewrite the expression as
![=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B4%5D%7B%282%5Ccdot%205%5E4%29%5E3%7D%3D5%5E3%20%5Csqrt%5B4%5D%7B2%5E3%7D%3D125%5Csqrt%5B4%5D%7B8%7D)
The calculated mutual inductance is 8.544 x 10⁻⁵ H.
Two coils have a mutual inductance of 1 henry when emf of 1 volt is induced in coil 1 and when the current flowing through coil 2 is changing at the rate of one ampere per second.
Length of the solenoid= 5.0 cm
Area of cross-section=1.0 cm²
no of spaced turns=300 turns
turns of insulated wire=180 turns
Mutual inductance (M) = μ₀μr N1N2 A/ L
=(4xπx 10⁻⁷) x (6.3 x 10⁻³) x 300 x 180 x 1/ 5
=79.12 x 10⁻¹⁰ x 54000 / 5
=8.544 x 10⁻⁵ H
hence, the mutual inductance is 8.544 x 10⁻⁵ H.
Learn more about Mutual inductance here-
brainly.com/question/14014588
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