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kolbaska11 [484]
3 years ago
5

Which of the following is the most likely end for a star that is small to average in size? A. blue main sequence B. red supergia

nt C. supernova D. white dwarf
Physics
2 answers:
likoan [24]3 years ago
7 0
"White dwarf" is the one among the following choices given in the question that is <span>the most likely end for a star that is small to average in size. The correct option among all the options that are given in the question is the last option or option "D". I hope that this is the answer that has actually helped you.</span>
stiv31 [10]3 years ago
3 0

Answer: D. White Dwarf

Explanation:

A star has life-cycle similar to any living specie. A star takes birth, goes through main sequence and dies in the end. The lifetime and the end of the star depend on the mass it holds.

A massive star  turns into red super-giant and goes supernova and ends up as either a neutron star or a black-hole.

On the other hand, a small to average star like our Sun will become a red giant and throw extra mass before ending up as a white dwarf.

The color on the main sequence (blue, yellow or red) depends on the temperature and age of the star. A young hot star is blue main sequence star. On the other hand, an old low temperature star towards the end of its life is a red star.

Hence, end for a star that is small to average in size is most likely to be a D. white dwarf.


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Alex_Xolod [135]

Answer:

Option C

Crimp terminals

Explanation:

It's possible to crimp terminals using a multipurpose wiring tool. Since the tool  selected for use during crimping also depends on the volume of work, the multipurpose wiring tool is recommended for use when the volume is small to medium. Basically, crimping tools are sized according to the wire gauge that they can fit. Since multipurpose has different sizes, that's why it's used for crimping tools.

5 0
4 years ago
30 points must be a legitimate answer and correct answer or I will report
kupik [55]

Answer:

it's C

Explanation:

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6 0
3 years ago
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From the list below, which name appears in two different parts of the body? *
Kisachek [45]

Answer:

Ngjfibfubbb

Explanation:

7 0
3 years ago
Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0
3 years ago
A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision
Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s

<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

<u>Average acceleration</u>

F=ma, so a=\frac{F}{m_{b}}.

Therefore, a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}

8 0
3 years ago
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