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Ratling [72]
2 years ago
9

The graph represents the change in that occurs when food is cooked over a charcoal grill. Which statement correctly explains the

graph?
A. The reactants are unlit charcoal that has already released its energy, and the products are charcoal that has already burned.
B. The reactants are charcoal that has already burned and released its energy, and the products are unlit charcoal.
C. The reactants are unlit charcoal, and the products are charcoal that has already burned and released its energy.
D. The reactants are charcoal that has already burned, and the products are unlit charcoal that has already released its energy.

Physics
1 answer:
MArishka [77]2 years ago
6 0

<u>O</u><u>p</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>C</u><u> </u><u>i</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>

<h3 /><h3><em>S</em><em>m</em><em>a</em><em>l</em><em>l</em><em> </em><em>Explanation</em><em>:</em><em>-</em></h3>

The reactants are charcoal that is unlit + oxygen and the products are the burnt charcoal + energy.

(Explanation with formula and reason attached. Check it.)

> Benjemin360

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Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i
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Answer:

The final speed for the mass 2m is v_{2y}=-1,51\ v_{i} and the final speed for the mass 9m is v_{1f} =0,85\ v_{i}.

The angle at which the particle 9m is scattered is \theta = -66,68^{o} with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

\vec{p}=m\vec{v}

p_{xi} =p_{xf}

In the x axis before the collision we have

p_{xi}=9m\ v_{i} - 2m\ v_{i}

and after the collision we have that

p_{xf} =9m\ v_{1x}

In the y axis before the collision p_{yi} =0

after the collision we have that

p_{yf} =9m\ v_{1y} - 2m\ v_{2y}

so

p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}

then

p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}

<u>Conservation of kinetic energy:</u>

\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}

so

\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}

Putting in one side of the equation each speed we get

\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the  speed of the 9m particle:

v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}

the magnitude of the final speed of the particle 9m is

v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }

v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}

The tangent that the speed of the particle 9m makes with the -y axis is

tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}

As a vector the speed of the particle 9m is:

\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}

As a vector the speed of the particle 2m is:

\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}

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