An astronomer examines an emission line spectrum for a star from data recorded by a spectroscope. There are four lines: two in the blue region, one in the green region, and one in the red region of the spectrum. How does the spectrum obtained from the spectroscope help the astronomer?

Answer:

Atoms can be joined together by chemical bonds.

Explanation:

When atoms form bonds, they can achieve a stable electron arrangement. To achieve a stable electron arrangement atoms can lose, gain or share electrons. There are many different types of bonds that hold atoms together.

Answer:

Atoms come together to form molecules because of their electrons. Electrons can join (or bond) atoms together in two main ways. When two atoms share electrons between them, they are locked together (bonded) by that sharing. These are called covalent bonds.

Explanation:

By looking at the model shown in the problem one can clearly say that this model represents the metal because there are a total of 12 electrons revolving around the nucleus which clearly represents the Magnesium metal.

The elements of the periodic tables that behave as metals, as well as the nonmetal in some chemical or physical aspects, are known as metalloids. Some examples of metalloids are Boron, silicon, germanium, arsenic, antimony, tellurium, etc.

As given in the problem, there is a graphical representation of the model of an atom,

One can see that the model in the problem accurately depicts the metal by looking at the total of 12 electrons circulating around the nucleus, which is a clear representation of the magnesium metal.

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The fraction of the kinetic energy of the ball lost during the collision is [tex]\frac{1}{3}[/tex].

The given parameters;

The initial and final momentum of the ball is calculated as;

[tex]P_i = m_ivi[/tex]

[tex]P_f = m_fv_f = \frac{1}{3} m_iv_i[/tex]

The initial and final kinetic energy of the ball is calculated as;

[tex]K.E_i = \frac{1}{2} m_iv_i^2 = \frac{1}{2} P_iv_i\\\\K.E_f = \frac{1}{2} m_fv_f^2= \frac{1}{2} (\frac{1}{3} P_iv_i)= \frac{1}{6} P_iv_i[/tex]

The change in the kinetic energy is calculated as;

[tex]\Delta K.E = K.E_f - K.E_i \\\\\Delta K.E = \frac{1}{6} P_iv_i - \frac{1}{2} P_iv_i = \frac{1}{3} P_iv_i[/tex]

Thus, the fraction of the kinetic energy of the ball lost during the collision is [tex]\frac{1}{3}[/tex].

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Answer:

a. 6

b. 6 m/s²

c. 300 m to the right

d. 30 secs

Explanation:

slope = rise /run

60-0/10-0

= 6

b. slope = acceleration = 6 m/s²

c. d=ut+1/2at²

t=10 (segment A last for 10 secs)

u - initial velocity = 0

so d = 0(10)+1/2*6*10²

=300 m

Answer:

-34.8n

Explanation:

when object in free fall, only force is the Force by weight,

which is 34.8 and direction is down.

Answer:

potentiometer is a device used to measure potential difference across the circuit

Answer:

its a sure for not but i think its a 192583

Answer:

it prevents slipping on the grass while playing

Answer:

to keep their grip on the ground so they won't fall

Answer:

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants. The law of conservation of mass is useful for a number of calculations and can be used to solve for unknown masses, such the amount of gas consumed or produced during a reaction.

Explanation:

<3

Answer:

3

Explanation:

It is part of the Earth's mantle

Answer:

it is the astenospere number 3

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

The vertical pressure of the water is calculated as follows;

[tex]P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\[/tex]

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10  ft

[tex]P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI[/tex]

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

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Answer:

B

Explanation:

I don't know but its b

Answer:

39.240 W

Explanation:

Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is [tex]w= \vec F\cdot \vec{h} = mg h = 1000 \times 9.81\times 40 = 392.400 J[/tex]

At this point we can simply apply the definition of power, that is [tex]P = \frac wt[/tex], to get the power of the engine is [tex]39.240 W[/tex]

Answer:

P=39200W

Explanation:

P=(mgh)/t

P=(40×9.8×1000)/10

P=39200W

The work done on the given gas in one cycle is -405.3 kJ.

The given parameters;

Convert the pressure to Pascal (N/m²);

1 atm = 101325

The work done in one cycle is the area of the triangle and it is calculated as follows

[tex]Area = \frac{1}{2} \times base \times height\\\\Area = \frac{1}{2} \times (3 \ m^3\ -\ 1 \ m^3)\times (6 \ atm \ - \ 2 \ atm)\\\\Area = 4 \ atm -m^3\\\\Area = 4 \ atm -m^3 \ \times \frac{101325 \ Pa}{1 \ atm} \\\\Area = 405,300 \ m^3.Pa\\\\Area = 405,300 \ m^3. (N/m^2)\\\\Area = 405,300 \ Nm\\\\Area = 405,300 \ J\\\\Area = 405.3 \ kJ[/tex]

the net work done on the gas = - 405.3 kJ

Thus, the work done on the given gas in one cycle is -405.3 kJ.

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Answer:

After solid particles condensed from the gas but before the planets finished forming

Answer:

The answer is False

Answer:

Conduction

Explanation:

Because when a molecule is transferred to another molecule then this in going into direct contact

For example if we have a pan and it is boiling well the heat that is on the pan holder that stays in place and it won`t move anywhere else like there is Radiation and Convection but for this question the answer would be Conduction

Answer:

Explanation:

A wind is the horizontal movement of air from an area of high pressure to an area of lower pressure. Winds are caused by differences in air pressure. Most differences in air pressure are caused by unequal heating of the atmosphere

Answer:

We have that the the minimum coefficient of static friction  is mathematically given as

[tex]\mu= 0.235[/tex]

Generally the equation for the Force   is mathematically given as

[tex]F=n\mu\\\\Where\\\\n = mg\\\\Given\\\\Fc = m(vt2/R)\\\\Therefore\\\\\mu = vt2/gR\\\\\mu = 2.22/(9.8(2.1)) \\\\[/tex]

[tex]\mu= 0.235[/tex]

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Answer:

It could be low self-esteem

Explanation:

There probably insecure have problems and issues of their own could be that were abused growing up and weren’t treated right who knows.

Answer:

H = V t - 1/2 g t^2         V and g are in different directions

-8.2 = V * 6.2 - 1/2 * 10 * 6.2*2 = 6.2 V - 192.2

V = 184 / 6.2 = 29.7 m/s

Check: find time for ball to return to zero height

0 = 29.7 T - 5 T^2

T = 29.7 / 5 = 5.94 sec

The ball must have fallen 8.2 m in (6.2 - 5.94) sec = .26 sec

S = 29.7 * .26 + 5 * .26^2 = 8.1

Using visual inspection and knowledge of linear regression , the inconsistent input is the last data point (30, 12.0).

Therefore, the inconsistent data point is (30, 12.0)

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Answer:

determine a large mam I have no idea

Answer:

time

Explanation:

power = work done ÷ time

Answer:

Explanation:

Use simplified formula:

Answer:

unfortunately we have not received a reply to the park yet but

Explanation:

mohamedz and he had killed him

Answer:

Approximately [tex]106\; \rm m[/tex], [tex](1/16)[/tex] of the original distance, assuming that the acceleration is constant.

Explanation:

If the acceleration on this vehicle is constant, the SUVAT equations would apply:

[tex]\displaystyle v^{2} - u^{2} = 2\, a\, x[/tex].

Rearrange this equation to find an expression for [tex]x[/tex], the displacement required for the vehicle to come to a stop:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{-u^{2}}{2\, a} && (\text{$v = 0\; \rm m \cdot s^{-1}$})\end{aligned}[/tex].

Thus, under these assumptions, [tex]x[/tex] would be proportional to [tex]u^{2}[/tex]. In other words, the distance required for this vehicle to come to a stop would be proportional to the square of the initial velocity of this vehicle.

If the initial velocity [tex]u[/tex] is reduced to [tex](1/4)[/tex] (a quarter) of the initial value, the distance required for stopping this vehicle would be [tex](1/4)^{2} = (1/16)[/tex] of the initial value:

[tex]\begin{aligned}1,\!697\; {\rm m} \times \frac{1}{16} \approx 106\; \rm m\end{aligned}[/tex].