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3 years ago

The electric field associated with a uniformly charged hollow metallic sphere is the greatest at:

1 answer:

3 0

<span>Electric field is a region around a charge in which the force of the charge can be exerted on the other charged particle. The force can be both inwards and outwards (repelling). Usually when it comes to a uniformly charged hollow metallic sphere the electric field associated is highest at outer surface of the sphere.</span>

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The surface temperature of a planet depends on both the distance of the planet from the Sun and on how the panet's atmosphere di

BaLLatris [955]

Answer:

Aphelion: 6404 W/m2

Perihelion: 14978 W/m2

Explanation:

The solar energy flux depends on the solar power output divided by the surface of a sphere with a radius equal to the distance to the Sun.

$\Phi sol = \frac{Psol}{4 * \pi * d^2}$

The distances we need are the aphelion and perihelion of Mercury.

Planetary orbits are ellipses. In an ellipse the eccentricity is related to linear eccentricity and the length of the semi major axis:

$e = \frac{c}{a}$

Where

e: eccentricity

c: linear eccentricity

a: semi major axis

The linear eccentricity is equal to the distance of the focus of the center of the ellipse.

$c = a * c =$

a = 0.39 AU = 5.83e10 m

$c = 5.83e10e * 0.21 = 1.22e10 m$

In planetary orbits the Sun is in one of the fucuses. With this we can calculate the prihelion and aphelion as:

Ap = a + c = 5.83e10 + 1.22e10 = 7.05e10 m

Pe = a - c = 5.83e10 - 1.22e10 = 4.61e10 m

And the solar energy fluxes will be:

$\Phi Ap = \frac{4e26}{4 * \pi * 7.05e10^2} = 6404 W/m2$

$\Phi Pe = \frac{4e26}{4 * \pi * 4.61e10^2} = 14978 W/m2$

4 0

3 years ago

Graph the following data on the graph, then use the graph to determine the half-life of this isotope.

Harlamova29_29 [7]

Answer:

4, 56

Explanation:

Hope this helped!

7 0

3 years ago

Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the

Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) , is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0

4 years ago

A car moved 50 km to the North. What is its displacement?

MariettaO [177]

Answer:

50 Km

Explanation:

If the car moved 50 km north and then 50 south its displacement would be 0. if it go in a direction like sw or ne or nw

4 0

4 years ago

PLEASE HELP ME! IM TIMED

vlada-n [284]

Answer:

Toothpaste

Because it is viscous

8 0

3 years ago