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vesna_86 [32]
3 years ago
15

The focal length for a spherical convex mirror is –20 cm. What is its radius of curvature?

Physics
1 answer:
Svetradugi [14.3K]3 years ago
4 0

Answer:

40 cm

Explanation:

The focus of a spherical convex mirror is the point at which the rays of an object converge in the infinite, it is also the point at which an object must be placed so that its image is formed in the infinite. The distance from the focus to the origin is called the focal length and is called f. It is related to the radius of the mirror, R, according to:

f=-\frac{R}{2}

rewriting for R:

R=-2f=-2*-20cm=40 cm.

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What happens if the air pressure in your throat and outside body is less than the air pressure in your middle ear when you swall
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Answer:

When the air pressure in the throat and outside the body is less than the air pressure in the middle ear, barotrauma occurs.

Explanation:

Ear barotrauma is a medical condition that describes discomfort in the ear which is caused by pressure differences in the inner and outer ear drum.

Usually, the air pressure in the middle ear is the same as the air pressure in the throat and outside the body.

When we swallow, the eustachian tube opens up and air flows out of and into the middle ear, this balances the pressure. But if the eustachian tube is blocked, the air pressure in the throat and outer body become different from the air pressure in the middle ear.

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Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for
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Calm, sunny days with wind moving away from the center.

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

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2 years ago
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If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum h
nadezda [96]

Answer:

H = 6.93 m

Explanation:

given data

velocity v = 35 m/s

horizontal component Vx = 33 m/s

solution

we get here maximum height so first we get vertical component here that is express as

Vy = \sqrt{v^2- Vx^2}        .........................1

put here value

Vy = \sqrt{35^2- 33^2}

Vy = 11.66 m/s

and

now we get height

H = \frac{Vy^2}{2g}        .............................2

put here value

H = \frac{11.66^2}{2\times 9.8}

H = 6.93 m

7 0
2 years ago
A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler
Pepsi [2]

Answer:

331665750000\ m/s^2

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = 1.989\times 10^{30}\ kg

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2

The gravitational acceleration at the surface of such a star is 331665750000\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s

The velocity of the object would be 3257806.62409 m/s

8 0
3 years ago
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