All categories

2 years ago

What kind of light would be the best to use to look inside a cold dark cloud and see the warm stars forming inside?

Physics

1 answer:

lys-0071 [83]2 years ago

7 0

<h2>Answer: Infrared light</h2>

A dark nebula is a cloud of dust and cold gas, which does not emit visible light and hides the stars it contains.

These types of nebulae are composed mainly of the hydrogen they obtain from nearby stars, which is their fuel.

It is using infrared light that we can "observe" and analyze in detail what happens in the inner parts of these nebulae.

You might be interested in

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

8 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

If you can't throw a football what do u need to work on more

NeX [460]

Work out your arms.

3 0

3 years ago

Read 2 more answers

In Milgram's experiment:

SpyIntel [72]

Answer:

B. The "Learner" was working with Milgram.

Explanation:

just took the test

give brainliest, please. :)

3 0

2 years ago

A 9.0-V battery (with nonzero resistance) and switch are connected in series across the primary coil of a transformer. The secon

qwelly [4]

Answer:

N₂ / N₁ = 13.3

Explanation:

A transformer is a system that induces a voltage in the secondary due to the variation of voltage in the primary, the ratio of voltages is determined by the expression

ΔV₂ = N₂ /N₁ ΔV₁

where ΔV₂ and ΔV₁ are the voltage in the secondary and primary respectively and N is the number of windings on each side.

In this case, they indicate that the primary voltage is 9.0 V and the secondary voltage is 120 V

therefore we calculate the winding ratio

ΔV₂ /ΔV₁ = N₂ / N₁

N₂ / N₁ = 120/9

N₂ / N₁ = 13.3

s good clarify that in transformers the voltage must be alternating (AC)

3 0

3 years ago