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2 years ago

Would you rather be beautiful but you cant look at yourself or be ugly and no one else can see you

1 answer:

6 0

Whats the point of being either of them?

If your Beautiful, you cant look at yourself...

And if your ugly, no one else can see you...

I'd rather not exist LOL XD

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The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t

Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = $854\times 10^{3}\ Hz$

Power, P = 50 kW = $50\times 10^{3}\ W$

$I_{p} = 1\ photon/s/m^{2}$

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

$n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s$

Area of the sphere, A = $4\pi r^{2}$

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is $1\ photon/s/m^{2}$

$I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}$

$1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}$

$r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m$

Now,

We know that:

1 ly = $9.4607\times 10^{15}\ m$

Thus

$r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly$

5 0

2 years ago

A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the

Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$ . What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1$

$x=2$

and $\rho = 2 \ kg/m^2$ , $P_x=1 \$ , $P_y=-2 \$ .

So,

$dI = dmr^2$

$dI = \rho \ dA \ r^2$ , $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

$= 3050.19 \ kg \ m^2$

So the moment of inertia is $3050.19 \ kg \ m^2$ .

4 0

2 years ago

Plz help need answer ASAP thanks

MAVERICK [17]

It’s procedural memory Bc idk but procedural is how to do something and declarative is remembering something

3 0

2 years ago

Statistically, consumers tell more people about negative experiences in business encounters than about positive experiences. A)

elena-s [515]

true hope this helps

8 0

2 years ago

Read 2 more answers

A particle has a charge of -4.25 nC.

SpyIntel [72]

Answer:

-611.32 N/C

0.43723 m

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = -4.25 nC

r = Distance from particle = 0.25 m

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C$

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

$E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m$

The distance from the electric field is 1.71436 m

4 0

2 years ago

Read 2 more answers