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3 years ago

I need help with this physics question

Physics

1 answer:

insens350 [35]3 years ago

4 0

It's impossible to describe WHERE a place is without mentioning ANOTHER place.

... Across the street from -- the bank.
... Next door to -- my house.
... 30 miles west of -- Chicago.
... Up above -- the tree.
... Two days ride out of -- Tulsa.
... Halfway home from -- school.
... Twice as far from Earth as -- the moon is.
... The first seat in -- the second row.
... Behind -- the dog's left ear.
... At the bottom of -- the pool.
... On the tip of -- my tongue.
... In the front seat of -- the car.
... I saw you in -- my dream.
... You're always on -- my mind.

The question is trying to get you to realize that to get from a reference point to a certain position, you have to know

How far
and
In what direction.

You might be interested in

An electron moving to the left at 0.8c collides with a photon moving to the right. After the collision, the electron is moving t

SVETLANKA909090 [29]

Answer:

Wavelength = 2.91 x 10⁻¹² m, Energy = 6.8 x 10⁻¹⁴

Explanation:

In order to show that a free electron can’t completely absorb a photon, the equation for relativistic energy and momentum will be needed, along the equation for the energy and momentum of a photon. The conservation of energy and momentum will also be used.

E = y(u) mc²

Here c is the speed of light in vacuum and y(u) is the Lorentz factor

y(u) = 1/√[1-(u/c)²], where u is the velocity of the particle

The relativistic momentum p of an object of mass m and velocity u is given by

p = y(u)mu

Here y(u) being the Lorentz factor

The energy E of a photon of wavelength λ is

E = hc/λ, where h is the Planck’s constant 6.6 x 10⁻³⁴ J.s and c being the speed of light in vacuum 3 x 108m/s

The momentum p of a photon of wavelenght λ is,

P = h/λ

If the electron is moving, it will start the interaction with some momentum and energy already. Momentum of the electron and photon in the initial and final state is

p(pi) + p(ei) = p(pf) + p(ef), equation 1, where p refers to momentum and the e and p in the brackets refer to proton and electron respectively

The momentum of the photon in the initial state is,

p(pi) = h/λ(i)

The momentum of the electron in the initial state is,

p(ei) = y(i)mu(i)

The momentum of the electron in the final state is

p(ef) = y(f)mu(f)

Since the electron starts off going in the negative direction, that momentum will be negative, along with the photon’s momentum after the collision

Rearranging the equation 1 , we get

p(pi) – p(ei) = -p(pf) +p(ef)

Substitute h/λ(i) for p(pi) , h/λ(f) for p(pf) , y(i)mu(i) for p(ei), y(f)mu(f) for p(ef) in the equation 1 and solve

h/λ(i) – y(i)mu(i) = -h/λ(f) – y(f)mu(f), equation 2

Next write out the energy conservation equation and expand it

E(pi) + E(ei) = E(pf) + E(ei)

Kinetic energy of the electron and photon in the initial state is

E(p) + E(ei) = E(ef), equation 3

The energy of the electron in the initial state is

E(pi) = hc/λ(i)

The energy of the electron in the final state is

E(pf) = hc/λ(f)

Energy of the photon in the initial state is

E(ei) = y(i)mc2, where y(i) is the frequency of the photon int the initial state

Energy of the electron in the final state is

E(ef) = y(f)mc2

Substitute hc/λ(i) for E(pi), hc/λ(f) for E(pf), y(i)mc² for E(ei) and y(f)mc² for E(ef) in equation 3

Hc/λ(i) + y(i)mc² = hc/λ(f) + y(f)mc², equation 4

Solve the equation for h/λ(f)

h/λ(i) + y(i)mc = h/λ(f) + y(f)mc

h/λ(f) = h/lmda(i) + (y(i) – y(f)c)m

Substitute h/λ(i) + (y(i) – y(f)c)m for h/λ(f) in equation 2 and solve

h/λ(i) -y(i)mu(i) = -h/λ(f) + y(f)mu(f)

h/λ(i) -y(i)mu(i) = -h/λ(i) + (y(f) – y(i))mc + y(f)mu(f)

Rearrange to get all λ(i) terms on one side, we get

2h/λ(i) = m[y(i)u(i) +y(f)u(f) + (y(f) – y(i)c)]

λ(i) = 2h/[m{y(i)u(i) + y(f)u(f) + (y(f) – y(i))c}]

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

Calculate the Lorentz factor using u(i) = 0.8c for y(i) and u(i) = 0.6c for y(f)

y(i) = 1/[√[1 – (0.8c/c)²] = 5/3

y(f) = 1/√[1 – (0.6c/c)²] = 1.25

Substitute 6.63 x 10⁻³⁴ J.s for h, 0.511eV/c2 = 9.11 x 10⁻³¹ kg for m, 5/3 for y(i), 0.8c for u(i), 1.25 for y(f), 0.6c for u(f), and 3 x 10⁸ m/s for c in the equation derived for λ(i)

λ(i) = 2h/[m.c{y(i)(u(i)/c) + y(f)(u(f)/c) + (y(f) – y(i))}]

λ(i) = 2(6.63 x 10-34)/[(9.11 x 10-31)(3 x 108){(5/3)(0.8) + (1.25)(0.6) + ((1.25) – (5/3))}]

λ(i) = 2.91 x 10⁻¹² m

So, the initial wavelength of the photon was 2.91 x 10-12 m

Energy of the incoming photon is

E(pi) = hc/λ(i)

E(pi) = (6.63 x 10⁻³⁴)(3 x 10⁸)/(2.911 x 10⁻¹²) = 6.833 x 10⁻¹⁴ = 6.8 x 10⁻¹⁴

So the energy of the photon is 6.8 x 10⁻¹⁴ J

6 0

2 years ago

The electron beam inside a television picture tube is 0.40 {\rm mm} in diameter and carries a current of 50 {\rm \mu A}. This el

Harrizon [31]

Answer:

A.3.13x10^14 electrons

B.330A/m²

C.9.11x10^5N/C

D. 0.23W

.pls see attached file for explanations

8 0

2 years ago

The mass of an object on earth is 20 kg what will be the mass of that object on moon? And why?

Alla [95]

Answer:

20kg

Explanation:

Mass is a measure of the amount of matter in an object. The mass of an object, the amount of matter inside it does not change based on location. E.g. Objects do not lose matter when they travel to the moon.

Weight, on the other hand is the downward force you exert on the ground. Weight is calculated by multiplying the mass by the gravitational field strength and changes in different places with different gravitational strength. E.g. The moon's gravitational strength is 1/5 of Earth's so the mass of the object would stay the same but the weight would be only 20% of the weight is had on earth.

Hope this helped!

7 0

2 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$