Ask question

Ask question

All categories

3 years ago

6

X-rays cannot pass through Earth's atmosphere. Which of these is the best location to place a telescope used to observe x-rays f

rom stars?
mountain
top of building
space
high-flying aircraft

2 answers:

Amiraneli [1.4K]3 years ago

7 0

Mountains, tops of buildings, and high-flying aircraft are all part of Earth's atmosphere, no matter how high they are. On the other hand, space doesn't belong to our atmosphere, it is outside of it. Having this in mind, the best location to place a telescope used to observe x-rays from stars is in space.

VashaNatasha [74]3 years ago

4 0

Space because you are closest to the stars

You might be interested in

Three kilograms of argon (Ar) changes from an initial volume and a temperature of 298K to (a) four times the volume and a temper

Kay [80]

Answer:

Explanation:

3 kg = 3 / 40 = .075 moles = n

a ) Since the gas is expanding isothermally ( temperature being constant )

work done by the gas

= 2.303 n RT log V₂ / V₁

Here V₂ / V₁ = 4 , T = 298

Put these values in the equation above ,

work done = .075x 2.303 x 8.312 x 298 log 4

= 257.6 J

b) In adibatic change

$pv^\gamma = constant$

$T V^{\gamma-1} = constant$

T₁ / T₂ = $(\frac{V_2}{V_1} )^{\gamma-1}$

298 / T₂ = $( 4)^{1.667-1}$

T₂ = 751.26 K.

In adiabatic change work done

= n R ( T₁ - T₂) / (γ -1)

.075x 8.312 X ( 751 - 298 ) / .667

= 423.38. J

8 0

3 years ago

ZanzabumX [31]

Answer:

a) El valor de la densidad es 0.79 $\frac{g}{cm^{3} }$ o 790 $\frac{g}{cm^{3} }$

b) El peso especifico es 7749.9 $\frac{N}{m^{3} }$

Explanation:

a) La densidad se define como la propiedad que tiene la materia, ya sean sólidos, líquidos o gases, para comprimirse en un espacio determinado. En otras palabras, la densidad es una magnitud que permite medir la cantidad de masa que hay en determinado volumen de una sustancia. Entonces, la expresión para el cálculo de la densidad es el cociente entre la masa de un cuerpo y el volumen que ocupa:

$d=\frac{m}{V}$

En este caso:

masa= 237 g= 0,237 kg (siendo 1000 g= 1 kg)
volumen= 300 cm³= 0,0003 m³ (siendo 1 cm³= 0,000001 m³)

Reemplazando:

$d=\frac{237 g}{300cm^{3} }$ → d=0.79 $\frac{g}{cm^{3} }$

$d=\frac{0,237 kg}{0,0003m^{3} }$ → d=790 $\frac{g}{cm^{3} }$

<u><em>El valor de la densidad es 0.79 </em></u> $\frac{g}{cm^{3} }$ <u><em> o 790 </em></u> $\frac{g}{cm^{3} }$ <u><em></em></u>

b) El peso específico es la relación existente entre el peso y el volumen que ocupa una sustancia en el espacio.

Entonces, en este caso, siendo el peso:

P= m*g= 0,237 kg* 9,81 $\frac{m}{s^{2} }$ = 2,32497 N

el peso especifico es calculado como:

$Pe=\frac{Peso}{Volumen}= \frac{2,32497N}{0,0003 m^{3} }$

Pe= 7749.9 $\frac{N}{m^{3} }$

<u><em>El peso especifico es 7749.9</em></u> $\frac{N}{m^{3} }$ <u><em></em></u>

8 0

3 years ago

The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between t

Radda [10]

Answer:

a) $F = 6816.5680 N$

b) $F' = 0 N$

c) $F''=28195.5N$

Explanation:

Magnitude of charges $M=1.6\muC$

Distance $d=2.6$

Generally the equation for Net Force is mathematically given by

For First Drawing

$F =\frac{ k q^2}{d^2}+\frac{k q^2}{d^2}$

$F =\frac{2* 9*10^9* (1.6*10^-6)^2}{ (2.6*10^-3^2}$

$F = 6816.5680 N$

For second Drawing

$F' =\frac{ k q^2}{d^2 }-\frac{ k q^2}{ d^2}$

$F' = 0 N$

For Third Drawing

$F'' =\frac{ k q^2}{d^2} * \sqrt (2)$

$F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}$

$F''=28195.5N$

4 0

3 years ago

Which of the following is example of magnitude in a specific direction

k0ka [10]

1. Velocity and 2. Acceleration

8 0

3 years ago

how much current is in a circuit that includes a 9-volt battery and a bulb with a resistance of 3 ohms?

ExtremeBDS [4]

3 Amper 9 divided by 3

8 0

3 years ago