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3 years ago

How do waves behave differently to Earth's interior as they encounter the boundaries of different mediums?

Physics

1 answer:

Natalka [10]3 years ago

3 0

Answer:

Explanation:

When they encounter boundaries between different media, the waves react according to Snell’s law, and the angle of refraction across the boundary will depend on the velocity of the second media relative to the first

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Light with a wavelength of 395 nm illuminates a metal cathode. The maximum kinetic energy of the emitted electrons is 0.76 eV .

lutik1710 [3]

Answer:

520.8 nm

Explanation:

We are given that

$\lambda=395 nm=395\times 10^{-9} m$

1 nm= $10^{-9} m$

Maximum kinetic energy, $K_{max}=0.76 eV=0.76\times 1.6\times 10^{-19}$ V

$1 e=1.6\times 10^{-19} C$

We have to find the maximum wavelength of light.

We know that

$h\nu=\frac{hc}{\lambda_0}+K_{max}$

Where $c=3\times 10^8 m/s$

$h=6.625\times 10^{-34}$

$6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}=\frac{6.625\times 10^{-34}\times 3\times 10^8}{\lambda_0}+0.76\times 1.6\times10^{-19}$

$6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}- 0.76\times 1.6\times10^{-19}=\frac{6.625\times 10^{-34}\times 3\times 10^8}{\lambda_0}$

$\lambda_0=\frac{6.625\times 10^{-34}\times 3\times 10^8}{6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}- 0.76\times 1.6\times10^{-19}}$

$\lambda_0=5.208\times 10^{-7} m$ =520.8 nm

6 0

3 years ago

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release

Dmitry [639]

THIS IS THE COMPLETE QUESTION BELOW

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release the ball 1.1 m above the ground

And what speed does the ball hit the ground? Solve this problem using energy.

Answer

a)minimum speed must you toss a 130 g is 15.9090m/s

b)speed the ball hit the ground is 16.57m/s

Explanation:

a)We know that For any closed/isolated system, the total energy is CONSERVED.

K.E. lost by the ball=The change in P.E of the ball at 1.1m above the ground as well as the P.E. of the ball at 14 m-high roof

This statement can be expressed as the expression below from K.E and P.E energy formula

P.E. = mgh

K.E. = (1/2)mv^2

Therefore,

(mgh1 - mgh2)=(1/2)mv^2

Where h2=the ball height above the ground=1.1m

h1=ball height at roof of the gymnasium= 14m

Then if we substitute we have

[(10) x (0.14) x (9.81)] - [(1.1) x (0.14) x (9.81)] = (1/2)(0.14)(v^2)

16.45137=0.065V^2

V=15.9090m/s

minimum speed must you toss a 130 g is 15.9090m/s

b)To calculate the speed the ball hit the ground?

This is the highest point (14m-high roof),and the type of the energy the ball possesses is Po.tential energy only.

At the lowest point (ground), the energy the ball possesses is K.E. only.

P.E at 10m-high roof = K.E. at ground.

(14) x (0.13) x (9.81) = (1/2) x (0.13) x v^2

17.8542= 0.065V^2

V= 16.57

Therefore,And speed the ball hit the ground is 16.57m/s

6 0

3 years ago

the crust is the thinnest layer on Earth, which allowed it to have a more constant theory of its composition. how do you think s

OlgaM077 [116]

Answer:

Explanation:

Geologists record the seismic waves and study how they travel through Earth. The speed of these seismic waves and the paths they take reveal how the planet is put together. Using data from seismic waves, geologists have learned that Earth's interior is made up of several layers.

3 0

3 years ago

Input Voltage = 220V

muminat

Answer:

Correct option is

450 V and 15 A

⇒T

200N

3000W

=15A

Efficciencyofthetransformer

η=

out

⇒

100

3000

100×6

90×5000

=450V

Hence,

option (A) is correct answer.

4 0

2 years ago

What acceleration does the force of earth's gravity peoduce

maksim [4K]

Near Earth's surface, gravitational acceleration is approximately 9.81 m/s2, which means that, ignoring the effects of air resistance, the speed of an object falling freely will increase by about 9.81 metres per second every second.

8 0

3 years ago