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Masteriza [31]
2 years ago
11

A star is located at a distance of about 100 million light years from Earth. An astronomer plans to measure the distance of the

star by the parallax method. Which of these statements best explains why the astronomer's method of measuring the star distance is inappropriate?
a. At large distances, a red shift in the star's spectrum causes incorrect measurements.
b. At large distances, a blue shift in the star's spectrum causes incorrect measurements.
c. The parallaxes beyond a few thousand light years are too large to be used to measure distances.
d. The parallaxes beyond a few thousand light years are too small to be measured with common instruments.
Physics
1 answer:
nlexa [21]2 years ago
5 0
<span>d. The parallaxes beyond a few thousand light years are
too small to be measured with common instruments.

I'm not sure that parallax can even be used out to a few
thousand light years.

The NEAREST star to Earth has the BIGGEST parallax.
The star is Alpha Centauri.  It's only 4 light years away
from us, and its parallax is  0.000206 of a degree !
I have no idea how astronomers can measure angles
so small ... and that's the BIGGEST parallax angle of
ANY star.</span>
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ioda

Answer:

6.1875\times 10^{-8}

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies I=\frac {k}{r^{2}} where r is the distance

Since intensity is given then when at 3 m

1.1\times 10^{-7}= \frac {k}{3^{2}}

k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}

Since we have the constant then at 4m

Intensity, I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}

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1 year ago
A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface
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Explanation:

Electric field strength= Force/unit charge

E= (kQq/r²)/q ₓ r

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E= \frac{kQ}{r^2} r^

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2 years ago
Un niño de 25 kg corre por un jardín con una velocidad de 2.5 m/s de forma que su trayectoria es tangente al borde de un carruse
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Answer:

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

Explanation:

Asumamos que tanto el niño como el carrusel no tienen carga externa aplicada sobre aquellos, de modo que se puede aplicar el Principio de Conservación de la Cantidad de Movimiento Angular:

m\cdot v \cdot R = (m\cdot R^{2}+I)\cdot \omega (1)

Donde:

m - Masa del niño, medida en kilogramos.

v - Velocidad lineal inicial del niño, medida en metros por segundo.

R - Radio máximo del carrusel, medida en metros.

I - Momento de inercia del carrusel, medida en kilogramo-metros cuadrados.

\omega - Velocidad angular final del sistema niño-carrusel, medida en radianes por segundo.

Si sabemos que m = 25\,kg, v = 2.5\,\frac{m}{s}, R = 2\,m y I = 500\,kg\cdot m^{2}, tenemos que la velocidad angular final es:

\omega = \frac{m\cdot v\cdot R}{m\cdot R^{2}+I}

\omega = \frac{(25\,kg)\cdot \left(2.5\,\frac{m}{s} \right)\cdot (2\,m)}{(25\,kg)\cdot (2\,m)^{2}+500\,kg\cdot m^{2}}

\omega = 0.208\,\frac{rad}{s}

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

4 0
2 years ago
A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point
finlep [7]

Answer:

1.17894 m

Explanation:

The rock is at one end of the rod which is 0.211 m from the fulcrum

F = Force

d = Distance

L = Length of rod

M = Mass of rock = 325 kg

g = Acceleration due to gravity = 9.81 m/s²

Torque

\tau=F\times d

Torque of man

\tau_m=F(L-d)\\\Rightarrow \tau_m=695(L-0.211)

Torque of rock

\tau_r=Mg\times d\\\Rightarrow \tau=325\times 9.81\times 0.211\\\Rightarrow \tau=672.72075\ Nm

The torques acting on the system is conserved

\tau_m=\tau_r\\\Rightarrow 695(L-0.211)=672.72075\\\Rightarrow L-0.211=\frac{672.72075}{695}\\\Rightarrow L-0.211=0.96794\\\Rightarrow L=0.96794+0.211\\\Rightarrow L=1.17894\ m

The length of the rod is 1.17894 m

5 0
2 years ago
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