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3 years ago

9

Which series consists of radioactive nuclides produced by successive radioactive decay until a stable nuclide is reached?

1 answer:

Karolina [17]3 years ago

5 0

decay series consists of radioactive nuclides produced by successive radioactive decay until a stable nuclide is reached. (…)

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A certain fluid (with properties listed below) is heated from 25 to 75 °C as it moves at 0.2 m/s through a straight thin-walled

dexar [7]

Answer:

not sure. I'll try answering this later

Explanation:

I'm not sure. I'll try answering this later .

3 0

3 years ago

Compare and contrast tension,compression and extension

Semmy [17]

In physical science, tension can be described as the motion that applies a pulling force that is transmitted axially through cables, chains, or string. On the other hand, compression is the application of balanced inward forces while extension is the application of balanced outward forces in the same axis through the object.

8 0

3 years ago

A musician on a unicycle is riding at a steady 6.2 m/s while playing their well tuned oboe at an A above middle C (440 Hz). They

Verizon [17]

Answer:

The frequency of the beat is $f_t = 8Hz$

Explanation:

From the question we are told that

The speed of the ride $v = 6,2 m/s$

The frequency of the oboe is $f = 440Hz$

The temperature outside is $T = 27^oC$

Generally the speed of sound generated in air is mathematically evaluated as

$v_s = 331 + 0.61 T$

Substituting value

$v_s = 331 + 0.61 *27$

$v_s = 347.47 \ m/s$

The frequency of sound(generated by the musician on he park) getting to the musicians on the unicycle is mathematically evaluated as

$f_a = \frac{v_s }{v_s - v} f$

substituting values

$f_a = \frac{347.47 }{347.47 - 6.2} * 440$

$f_a = 448Hz$

Since the musician on the park is not moving the frequency of sound (from the musicians riding the unicycle )getting to him is = 440Hz

The beat frequency these musician here is mathematically evaluated as

$f_t = f_a - f$

So $f_t = 448 - 440$

$f_t = 8Hz$

3 0

3 years ago

a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric

n200080 [17]

Part a)

Here in this we can use momentum conservation as there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}$

here we know that

$m_1 = 0.004 kg$

$v_{1i} = 650 m/s$

$m_2 = 0.095$

$v_{2i} = 0$

$v_{2f} = 23 m/s$

now by above equation

$0.004*650 + 0.095* 0 = 0.004*v + 0.095*23$

$2.6 + 0 = 0.004*v + 2.185$

$v = 103.75 m/s$

Part b)

Final kinetic energy of the system

$KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

$KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2$

$KE_f = 46.65 J$

Initial Kinetic energy of the system will be

$KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2$

$KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2$

$KE_f = 845 J$

So here kinetic energy is decreased for this system

final energy is less than initial energy

6 0

3 years ago

When two forces or more forces act on an object, the forces are

sashaice [31]

The forces are net forces. I think this is correct.

5 0

3 years ago