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3 years ago

Define rotational speed?- short answer

Physics

1 answer:

Marina86 [1]3 years ago

3 0

Rotational speed of an object around an axis is the numbers of turns of the object divided by time. <span />

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Can someone please help me with this question? ASAP thank you!

Assoli18 [71]

did you get the answer

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2 years ago

Non examples of gravity ?

Afina-wow [57]

Floating. When you have no gravity you have nothing to be pushing you down to the floor so that would be an example of no gravity pushing on you.

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2 years ago

Read 2 more answers

The electromagnetic wave that delivers a cellular phone callto

spin [16.1K]

Answer:

$E=2.41\cdot 10^{-5} J$

Explanation:

The intensity of an electromagnetic wave can be expressed in terms of the magnetic field using the next relationship:

$I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}}$ (1)

c is the speed of light (3*10⁸ m/s)
μ₀ is the permeability of free space (in vacuum ) (1.26*10⁻⁶ N/A²)
B₀ is the magnetic field

$I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}$

$I_{average}=2.68\cdot 10^{-6} W/m^{2}$

Now, let's define the relationship between power (P) and average intensity (I).

$I_{average}=\frac{P}{A}$

P is the power
A is the area crossed

So we can calculate the power.

$P=I_{average}\cdot A=2.68\cdot 10^{-6}\cdot 0.20=5.37\cdot 10^{-7} W$

Finally, energy is the product of P times time, so:

$E=P\cdot t=5.37\cdot 10^{-7} \cdot 45=2.41\cdot 10^{-5} J$

I hope it helps you!

5 0

2 years ago

Large amounts of water that quickly infiltrate soil on a sloped surface can

sergey [27]

A is the correct choice

7 0

2 years ago

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long

joja [24]

Answer:

$5.65487\times 10^{-8}\ Wb$

$1.17\times 10^{-5}\ H$

-0.020475 V

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

$N_1$ = Number of turns of coil = 25

$N_2$ = Number of turns of coil 2 = 300

$\frac{di_2}{dt}$ = Rate of current increased = $1.75\times 10^3\ A/s$

d = Diameter = 2 cm

r = Radius = $\frac{d}{2}=\frac{2}{2}=1\ cm$

A = Area = $\pi r^2$

Magnetic field in the solenoid is given by

$B=\mu_0\frac{N_2}{l}I\\\Rightarrow B=4\pi\times 10^{-7}\frac{300}{0.25}\times 0.12\\\Rightarrow B=0.00018\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00018\times \pi\times 0.01^2\\\Rightarrow \phi=5.65487\times 10^{-8}\ Wb$

The average magnetic flux through each turn of the inner solenoid is $5.65487\times 10^{-8}\ Wb$

Mutual inductance is given by

$L=\frac{N_1\phi}{i_1}\\\Rightarrow L=\frac{25\times 5.65487\times 10^{-8}}{0.12}\\\Rightarrow L=1.17\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $1.17\times 10^{-5}\ H$

Induced emf is given by

$V=-L\frac{di_2}{dt}\\\Rightarrow V=-1.17\times 10^{-5}\times 1.75\times 10^3\\\Rightarrow V=-0.020475\ V$

The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.020475 V

5 0

3 years ago