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3 years ago

Radioactive isotopes of an atom are:

Physics

2 answers:

Len [333]3 years ago

8 0

Radioactive isotopes of an atom aren't as stable as you would think, so the correct answer is: C. Less Stable

(I can verify this because I took the test.)

Hope this helps! Have a wonderfully wonderful day!

Cheers mate!

Leto [7]3 years ago

6 0

Answer:

c.less stable

Explanation:

A radioactive isotope is an isotope of an atom that has an unstable nucleus. This generally happens when the nucleus has a much larger number of neutrons compared to the number of protons: in this case, the nucleus tends to decay, emitting part of the neutrons (but also protons or electrons, depending on the type of decay), becoming a lighter nucleus which is more stable. Therefore, a radioactive isotope is generally not very stable.

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Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.

Sati [7]

Answer:

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$ , $z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$ , $z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

$z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]$

Where angles are measured in radians and k represents an integer between $0$ and $n - 1$ .

The magnitude of the complex number is $27$ and the equivalent angular value is $1.817\pi$ . The set of cubic roots are, respectively:

k = 0

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -0.978 + i\cdot 2.836$

k = 1

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= -1.967 - i\cdot 2.265$

k = 2

$z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]$

$z^{\frac{1}{3} }= 2.945 - i\cdot 0.571$

5 0

3 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

7 0

2 years ago

Why are coal,nuclear,oil,and natural gas determined to be non-renewable types of energy resources

IrinaVladis [17]

Answer:

Once used as a energy source they cannot be charged/used again.

Explanation:

7 0

3 years ago

A worker drives a 0.562 kg spike into a rail tie with a 2.26 kg sledgehammer. The hammer hits the spike with a speed of 64.4 m/s

zubka84 [21]

Answer:

Explanation:

Given that,

Mass of sledge hammer;

Mh =2.26 kg

Hammer speed;

Vh = 64.4 m/s

The expression fot the kinetic energy of the hammer is,

K.E(hammer) = ½Mh•Vh²

K.E(hammer) = ½ × 2.26 × 64.4²

K.E ( hammer) = 4686.52 J

If one forth of the kinetic energy is converted into internal energy, then

ΔU = ¼ × K.E(hammer)

∆U = ¼ × 4686.52

∆U = 1171.63 J

Thus, the increase in total internal energy will be 1171.63 J.

4 0

3 years ago

Which statement below concerning the photoelectric effect is true? View Available Hint(s) Which statement below concerning the p

kakasveta [241]

There is a threshold frequency for each metal, and only light of a frequency higher than this threshold causes electrons to be emitted from the metal surface.

8 0

3 years ago