Hello I want to ask questions about physics

The Sun delivers an average power of 15.29 W/m2 to the top of Saturn's atmosphere. Find the magnitudes of vector E max and vecto

Elenna [48]

Answer:

$E=\sqrt{\frac{2u_E}{\epsilon_0}}=1.8*10^6N/C\\B=\sqrt{2\mu_0}=1.58*10^{-3}T$

Explanation:

The energy density of electromagnetic waves can be computed as

$u_{E}=\frac{1}{2}\epsilon_0E^2\\u_B=\frac{B^2}{2\mu_0}$

The power is the energy per unit of time. Hence we can take uE and Ub as 15.29J

By taking apart E and B we have

$E=\sqrt{\frac{2u_E}{\epsilon_0}}=1.8*10^6N/C\\B=\sqrt{2\mu_0}=1.58*10^{-3}T$

hope this helps!!

6 0

3 years ago

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5. Annie drags her little red wagon with a mass of 5.00 kg, up a hill that has an angle of

maxonik [38]

Answer: $25.08\ m/s^2$

Explanation:

Given

mass of wagon m=5 kg

elevation

$\theta =30.9^{\circ}\\F=150\ N$

Sin component of weight will oppose the applied force therefore we can write

$F-W\sin \theta=ma\\where\\W=weight(mg)\\a=acceleration$

$150-5\times 9.8\times \sin (30.09)^{\circ}=5\times a\\125.43=5a\\a=25.08\ m/s^2$

6 0

3 years ago

What are three examples of objects that float in water?<br><br> (Please don't copy and paste)

mars1129 [50]

Objects like apples, wood, and sponges are less dense than water. They will float. Many hollow things like empty bottles, balls, and balloons will also float.

7 0

3 years ago

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PLEASE HELP <br> IS THIS CORRECT?

Umnica [9.8K]

Answer:

yes!

Explanation:

7 0

3 years ago

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A long, thin solenoid has 390 turns per meter and a radius of 1.20 cm. The current in the solenoid is increasing at a uniform ra

gtnhenbr [62]

To solve this problem it is necessary to apply the concepts related to Faraday's law and the induced emf.

By definition the induced electromotive force is defined as

$\int E dl = -\frac{d\phi}{dt}$

$\int E dl = -(\frac{dB}{dt})A$

Where,

$\phi =$ Electric field

B = Magnetic Field

A = Area

At the theory the magnetic field is defined as,

$B = \mu_0 NI$

Where,

N = Number of loops

I = current

$\mu_0 =$ Permeability constant

We know also that the cross sectional area, is the area from a circle, and the length is equal to the perimeter then

A = \pi r^2

l = 2\pi r

Replacing at the previous equation we have that

$E (2\pi r) = \mu_0 n (\frac{di}{dt})(\pi R^2)$

Where,

R = Radius of the solenoid

r = The distance from the axis

Re-arrange to find the current in function of time,

$\frac{di}{dt} = \frac{Er}{\mu_0 NR^2}$

Replacing our values we have

$\frac{di}{dt} = \frac{(8.00*10^{-6})(0.0348)}{(4\pi*10^{-7})(390)(1.2*10^-2)^2}$

$\frac{di}{dt} = 3.94487A/s$

8 0

3 years ago