Volcanos create rich soil for farming true or false<br> (30 points)

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

lys-0071 [83]

<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.

7 0

3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

3 years ago

Which question cannot be answered through making measurements?

Margarita [4]

Answer: it would be A

Explanation: how are we to measure the air of a square mile

8 0

4 years ago

Is a glacier nothing more than a huge mineral flowing downhill?

NNADVOKAT [17]

Yes. A glacier is nothing more than a huge mineral flowing downhill.

4 0

4 years ago

PHYSICS HELP<br> PLEASE HELP ITS ABOUT ATWOOD MACHINES

m_a_m_a [10]

Answer:

7.23407 $\frac{m}{s^2}$

Explanation:

(I will not include units in calculations)

I'm assuming FBD's are already drawn, so I will work from there.

Let the 2.2kg block equal $m_2$ , and the 20kg block equal $m_1$ .

Summation equation for $m_2$ : $\sum F_x=F_t_2-(F_f+F_g_x)=m_2a$ , $\sum F_y=F_n-F_g_y=0$

Summation equation for $m_1$ : $\sum F_y=F_g-F_t_1=m_1a$

Torque Summation Equation: $\sum\tau=F_t_1*r-F_t_2*r=I\alpha$

Do some plugging in with the values given: $\sum\tau=F_t_1*r-F_t_2*r=.5Mr^2\alpha$

Replace $\alpha$ with $\frac{a}{r}$ , and cancel out the r's.

$\sum\tau=F_t_1-F_t_2=.5Ma$

This step is important: Rearrange the force summation equation to solve for each tension force.

$F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a$

Perform Substitution: $\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma$

Now, we need to find the friction force and the horizontal component of the force of gravity.

Note that $F_f=$ μ $F_n$

And based on our earlier summation equation: $F_n=F_g_y$

First, break $F_g$ into x and y components. $F_g_y=F_g\cos(\theta)$ , $F_g_x=F_g\sin(\theta)$

Perform substitution with this and the fact that $F_g=mg$ .

$\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma$

Solving for a, plugging in numbers yields an answer of 7.23407 $\frac{m}{s^2}$

6 0

3 years ago

Volcanos create rich soil for farming true or false (30 points)