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Luden [163]
3 years ago
5

Which is an advantage of using space observatories for space exploration?

Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0

It allows scientists to avoid the effects of light pollution.

dangina [55]3 years ago
7 0

Answer:

The advantage is to avoid the effects of light pollution.

Explanation:

There are several reasons why space observation is desirable in order to avoid problems that terrestrial observatories have. An example of this is that the space observatory does not suffer from the interference of city lights, avoiding light pollution. In addition, thermal turbulence (present in the atmosphere) also does not alter the telescope image.  The terrestrial atmosphere, having a spherical shape, acts as a lens that distorts images (a fact known as optical aberration), so the resolution of terrestrial images has a significant reduction. The atmosphere of our planet also absorbs an important portion of the electromagnetic spectrum (infrared and ultraviolet rays, for example), so some observations are virtually impossible to make.

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Para fabricar la bicicleta de un niño pequeño se tiene en cuenta que la fuerza que puede desarrollar es menor que la de un adult
Georgia [21]

Answer:

a) El piñón debe tener 20 dientes.

b) La bicicleta avanza aproximadamente 1,759 metros por cada pedaleada completa.

Explanation:

a) El plato es el engranaje más grande que forma parte del sistema de transmisión, acompañando a la cadena y el piñón integrado a la rueda trasera. Asumiendo que no existen pérdidas por fricción seca y que las condiciones de lubricación del sistema de transmisión son óptimas tal que las pérdidas de potencia son despreciables. Además, supongamos que la bicicleta viaja a velocidad constante, entonces tenemos la siguiente identidad mediante las definiciones de trabajo y potencia:

T_{P}\cdot \omega_{P} = T_{p}\cdot \omega_{p} (1)

Donde:

T_{P} - Torque del plato, en newton-metros.

T_{p} - Torque del piñón, en newton-metros.

\omega_{p} - Rapidez angular del piñón, en radianes por segundo.

\omega_{P} - Rapidez angular del plato, en radianes por segundo.

Sabiendo el hecho que tanto el plato y el piñón experimenta la misma velocidad tangencial, podemos simplificar (1) como sigue:

\frac{T_{P}}{R_{P}} = \frac{T_{p}}{R_{p}} (1b)

Puesto que el radio de cada elemento y el número de dientes son, por separado, directamente proporcionales al número de dientes, modificamos (1b) así y tenemos la siguiente identidad, la cual equivale a su vez a la razón de desarrollo:

\frac{T_{P}}{T_{p}} = \frac{N_{P}}{N_{p}} = \frac{R_{P}}{R_{p}} = \frac{\omega_{p}}{\omega_{P}} (1c)

Donde:

N_{p} - Número de dientes del piñón, sin unidad.

N_{P} - Número de dientes del plato, sin unidad.

Si tenemos que r = 1,4 y N_{P} = 28, entonces tenemos que el número de dientes del piñón es:

r = \frac{N_{P}}{N_{p}}

N_{p} = \frac{N_{P}}{r}

N_{p} = \frac{28}{1,4}

N_{p} = 20

El piñón debe tener 20 dientes.

b) De acuerdo con la relación de desarrollo, por cada revolución realizada por el plato, el piñón realiza 1,4 revoluciones. Entonces, el avance realizado por la rueda trasera (s), en metros, es igual al productor de la relación de desarrollo y la circunferencia de la rueda, es decir:

s = r\cdot 2\pi\cdot R (1)

Donde R es el radio de la rueda trasera, en metros.

Si conocemos que r = 1,4 y R = 0,2\,m, entonces el avance realizado por la rueda trasera es:

s = r\cdot 2\pi\cdot R

s = (1,4)\cdot (2\pi)\cdot (0,2\,m)

s \approx 1,759 \,m

La bicicleta avanza aproximadamente 1,759 metros por cada pedaleada completa.

3 0
3 years ago
A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is
kotykmax [81]

323.5 N is the tension in the cable.

Given

Mass of crate(M) = 175.5 kg

Mass of boom(m) = 94.7 kg

The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.

The Angle of the boom with the horizontal can be calculated by

tanθ = 5/10

θ = tan⁻¹(5/10) = 26.56°

Angle of the boom with horizontal is 26.56°

The angle of cable with horizontal can be calculated by

tan B = 4/10

B = tan⁻¹(4/10) = 21.80°

Angle of cable with horizontal is 21.80°

Taking moments of force about the point X

(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1

(175.5 × cos 26.56 + 94.7 × cos 26.56 )× 0.5 = T (sin(26.56 + 21.80) X 1

By calculating, we get

Tension(T) = 241.68/0.747

Tension(T) = 323.5 N

Hence, 323.5 N is the tension in the cable.

Learn more about Tension here brainly.com/question/24994188

#SPJ1

8 0
2 years ago
¿Cuál será la potencia de una bombilla conectada a una red de energía eléctrica de 440V, si la corriente que circula por el circ
goblinko [34]

Answer:

La potencia de la bombilla es de 1056 W.

Explanation:

La potencia de la bombilla se puede calcular usando la siguiente ecuación:

P = I*V

En donde:

P: es la potencia

I: es la corriente = 2,4 A

V: es la diferencia de potencial = 440 V

Entonces, la potencia es:

P = I*V = 2,4 A*440 V = 1056 W

Por lo tanto, la potencia de la bombilla es de 1056 W.

Espero que te sea de utilidad!

3 0
3 years ago
A 2-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents per kWh, wh
frozen [14]

Answer:

<h2> $1.50</h2>

Explanation:

Given data

power P= 2 kW

time t= 15 min to hours = 15/60= 1/4 h

cost of power consumption per kWh= 10 cent = $0.1

We are expected to compute the cost of operating the heater for 30 days

but let us computer the energy consumption for one day

Energy of heater  for one day= 2* 1/4 = 0.5 kWh

the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50

<u><em>Hence it will cost  $1.50 for 30 days operation</em></u>

4 0
3 years ago
What is one reason why it is very difficult to directly take a picture of an extrasolar planet?
Setler [38]

Answer:

Extrasolar planets are very dim light sources compared to their stars. At visible wavelengths, they generally have less than a millionth of the brightness of their parent star. It is extremely difficult to detect this type of dim light source, and in addition, the parent star has dazzling light that almost makes it impossible.

5 0
3 years ago
Read 2 more answers
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