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3 years ago

Who knows the egg drop challenge

1 answer:

7 0

I know. If you're asking for ideas, I once did one and it did not break. I just took a cardboard box, filled it with thins like packing peanuts, crumpled newspaper, and put the egg inside a few layers of socks in the middle, and taped the box tightly shut. Our science teacher dropped from the roof of the school, and there wasn't even a crack on mine!

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In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

3 years ago

Which statement is true of a glass lens that diverges light in air?

Arisa [49]

C it is uniformly thick

3 0

2 years ago

Read 2 more answers

A) Find the displacement for the given time intervals ,

Olegator [25]

Answer:

hahsususisjsbsb

Explanation:

vsgshzjdjdk

3 0

3 years ago

Read 2 more answers

An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro

valkas [14]

Answer:

Final velocity of electron, $v=6.45\times 10^5\ m/s$

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$ ........(1)

a is the acceleration, $a=\dfrac{F}{m}$

We know that electric force, F = qE

$a=\dfrac{qE}{m}$

Use above equation in equation (1) as:

$v^2=u^2+\dfrac{2qEs}{m}$

$v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m$

v = 647302.09 m/s

$v=6.45\times 10^5\ m/s$

So, the final velocity of the electron when it reaches point B is $6.45\times 10^5\ m/s$ . Hence, this is the required solution.

3 0

3 years ago

Suppose a blood vessel's radius is decreased to 87% of its original value by plaque deposits and the body compensates by increas

dexar [7]

Answer:

The pressure difference will increase by the factor of 1.75

Explanation:

For constant flow rate, coefficient of viscosity, length of the vessel and the pressure difference is inversely proportional to the fourth power of the radius of the blood vessel

Apply the principle of Poiseuille’s law.

Q = (P2 - P1)/R

Pls check the attached file for step by step solution of the question. It is submitted in this way as typing the equation may not be explanatory.

3 0

3 years ago