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2 years ago

A common isotope of Carbon has an atomic number of 6 and a mass number of 13. How many neutrons are in an atom of this isotope?

Physics

1 answer:

Yuki888 [10]2 years ago

6 0

Answer:

D. 13

Explanation:

For most elements other than hydrogen, isotopes are named for their mass number. For example, carbon atoms with the usual 6 neutrons have a mass number of 12 (6 protons + 6 neutrons = 12), so they are called carbon-12. Carbon atoms with 7 neutrons have an atomic mass of 13 (6 protons + 7 neutrons = 13).

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A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor

Allisa [31]

Answer:

the impedance of the circuit is 25.7 ohms.

Explanation:

It is given that,

Voltage, V = 50 volts

Frequency, f = 60 Hz

Resistance, R = 25 ohms

Capacitive resistance, $X_C=18\ ohms$

Inductive resistance, $X_L=24\ ohms$

We need to find the impedance of the circuit. It is given by :

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$Z=\sqrt{25^2+(24-18)^2}$

Z = 25.7 ohms

So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.

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3 years ago

What is a material that restricts the flow of electricity or thermal energy

Alina [70]

Answer:

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Explanation:

The mineral wool is a porous material that restricts thermal energy by trapping air in.

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3 years ago

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An object with a mass of 10 kg is accelerated upward at 2 m/sec2. What force is required?

telo118 [61]

Answer:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

Explanation:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

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3 years ago

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Taya2010 [7]

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take 56.9/1.137 = 50.043 mi/hr

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3 years ago

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An airplane traveling at one third the speed of sound (i.e., 114 m/s) emits a sound of frequency 3.72 kHz. At what frequency doe

nlexa [21]

Answer:

f'=5.58kHz

Explanation:

This is an example of the Doppler effect, the formula is:

$f'=\frac{(v+v_o)}{(v+v_s)}f$

Where f is the actual frequency, $f'$ is the observed frequency, $v$ is the velocity of the sound waves, $v_o$ the velocity of the observer (which is negative if the observer is moving away from the source) and $v_s$ the velocity of the source (which is negative if is moving towards the observer). For this problem:

$f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s$

$f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz$

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3 years ago