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2 years ago

Please help with some/all questions on this worksheet

Physics

1 answer:

Roman55 [17]2 years ago

5 0

A. Yes electricity is a fuel this is because protons are being displaced threw currents of electricity that are the used too charge a coil witch spins causing rotationional force And power too appropriate machine
B. Octane
C.one advantage is more power since octane is more combustable than electricity it makes gas or Diesel engines more powerful as too why there are no semi trucks being powered by electricity.
D. One disadvantage is effectively how well its fuel is spent gas is 69.78% less efficient

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It is B. false that an instrument used to detect a static electric charge is called an ammeter. It is actually called an electroscope. Ammeter measures current.

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What do you need to know to establish motion

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Answer:

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Explanation:

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2 years ago

Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir

strojnjashka [21]

Answer:

$1.475\times 10^{-13}\ C/m^3$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

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Electric flux through the top would be

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At the bottom

$120A$

Total flux through the volume

$\phi=120-110\\\Rightarrow \phi=10A$

Electric flux is given by

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Charge per volume is given by

$\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3$

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3 years ago

Normalize the equations

tatyana61 [14]

Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

$\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k$

Part b)

$\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1$

$\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}$

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3 years ago

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Volgvan

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