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2 years ago

13

What is a heterogeneous mixture and what is a homogeneous mixture?

2 answers:

yuradex [85]2 years ago

5 0

A heterogeneous mixture is a mixture that contains two or more distinct substances that you can see. You can see the different part if a heterogeneous mixture. An example of this is a salad. You can see all of the parts. A homogeneous mixture is a mixture that is uniform and you cannot see the different parts. It is still a mixture though. An example of that would be salt water. The water and salt are not chemically combined but you cannot see the salt AND water. It is just one solution.

Katena32 [7]2 years ago

3 0

Answer:

heterogeneous- Is not evenly mixed throughout.

homogeneous- Is evenly mixed throughout.

Explanation:

(APEX)

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Which statement about the effects of medium on the speed of a mechanical wave is true?

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C. A mechanical wave generally travels faster in gases than liquids.

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3 years ago

The carbon isotope 14C is used for carbon dating of archeological artifacts. 14C(mass 2.34×10−26kg) decays by the process known

Nookie1986 [14]

Answer:

2240.92365 m/s

Explanation:

$m_1$ = Mass of electron = $9.11\times 10^{−31}\ kg$

$v_1$ = Speed of electron = $5.7\times 10^7\ m/s$

$p_2$ = Neutrino has a momentum = $7.3\times 10^{-24}\ kg m/s$

M = total mass = $2.34\times 10^{-26}\ kg$

In the x axis as the momentum is conserved

$Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s$

In the y axis

$Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s$

The resultant velocity is

$R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s$

The recoil speed of the nucleus is 2240.92365 m/s

3 0

3 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

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