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katen-ka-za [31]

3 years ago

15

some galaxies have regions that are relatively blue other regions appear redder what does this varuation indicate about the diff

erences between these regions?

1 answer:

Vinvika [58]3 years ago

3 0

Answer:

Blue is hot while the red is cool.

Explanation:

Hot stars appear in blue whereas cooler stars appear to have red hues.

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A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

4 0

3 years ago

For a cosine function with amplitude A=0.75 and period T=10, what is y(4)?

mixer [17]

<span>2π/T = 2π/10 = π/5

y(x) = A sin (wx) = 0.75 sin (πx/5)

y(4) = 0.75 sin (4π/5) = 0.4408389392... ≈ 0.441</span><span>
</span>

8 0

2 years ago

Read 2 more answers

Researchers in the Antarctic measure the temperature to be -32°F. What is this temperature on the following scales?(a) the Celsi

Illusion [34]

Answer:

a) -35.6°C

b) 237.4 K

Explanation:

To convert temperature from degree celsius to degree fahrenheit, use the formula below:

$T_c=\frac{5}{9}(T_f-32)$

a) Therefore to convert -32°F to celsius, substitute it into the celsius

$\begin{gathered} T_c=\frac{5}{9}(-32-32) \\ \\ T_c=\frac{5}{9}(-64) \\ \\ T_c=-35.6^0C \end{gathered}$

b) To covert to the Kelvin scale, use the formula below

$\begin{gathered} T_k=T_c+273 \\ \\ T_k=-35.6+273 \\ \\ T_k=237.4K \\ \end{gathered}$

8 0

1 year ago

Hey! Can someone help with this question? Thx :)

Schach [20]

<span>A. Chemical energy to chemical energy</span>

4 0

3 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago