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3 years ago

9

When a skydiver is falling and the force of air resistance is equal to the force of gravity, what happens to the skydiver's moti

on?

1 answer:

Reil [10]3 years ago

6 0

Answer:

Terminal Velocity

Explanation:

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Two wooden boxes of equal mass but different density are held beneath the surface of a large container of water. Box A has a sma

Helga [31]

Answer:

<em>Which box has the greater acceleration?</em>

E. Box A

Explanation:

<em>The question is incomplete:</em>

<em>Which box has the greater acceleration?</em>

The bouyant force exerted by the water is equal in both boxes, because it depends on the volume displaced (that is the same for both boxes) and the density of the water.

But, the weight of each boxes is different, according to their density.

For the Box A the acceleration will be:

$m_aa_a=gV(\rho_w-\rho_a)\\\\\rho_aVa_a=gV(\rho_w-\rho_a)\\\\a_a=g\frac{(\rho_w-\rho_a)}{\rho_a}$

The same applies for the Box B:

$a_b=g\frac{(\rho_w-\rho_b)}{\rho_b}$

If we express the ratio of the accelerations, we have:

$a_a/a_b=\frac{(\rho_w-\rho_a)}{\rho_a}*\frac{\rho_b}{(\rho_w-\rho_b)}\\\\$

$a_a/a_b=\frac{(\rho_w-\rho_a)}{(\rho_w-\rho_b)} \frac{\rho_b}{\rho_a}$

We know that both densities are lower than water, because they accelerate upward to the surface when they are released (if they were more dense than water, they would sink more).

We will treat the densities as relative to water, so it becomes rho_w=1.

If we distribute the product, and know that the density of B is higher than the density of A, and both are higher than the product of the densities, we have:

$\rho_w=1\\\\\frac{a_a}{a_b}=\frac{(1-\rho_a)}{(1-\rho_b)} \frac{\rho_b}{\rho_a}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}\\\\\\\rho_b>\rho_a>\rho_a\rho_b>0\\\\\\\frac{a_a}{a_b}=\frac{\rho_b-\rho_a\rho_b}{\rho_a-\rho_a\rho_b}>1\\\\a_a>a_b$

The acceleration of A is higher than the acceleration of B.

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4 years ago

An imaginary line perpendicular to a reflecting surface is called _________.

n200080 [17]

<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)

So, Your Answer would be Option B

Hope this helps!</span>

5 0

3 years ago

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A biochemical process that liberates energy content of food for living organs

jok3333 [9.3K]

Hello!

The <span>biochemical process that liberates energy content of food for living organs is called catabolism.

Catabolism is a process in which big molecules like those in the food we eat are broken down into small ones that can be used by the living organism to obtain energy. The energy that is liberated in this process is stored in molecules of adenosine triphosphate (ATP). These molecules are used by cells to produce energy in the process called cellular respiration, catalyzed by the enzyme ATPase.

Have a nice day!
</span>

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3 years ago

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

s2008m [1.1K]

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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3 years ago

Which can have either a positive charge or a negative charge when they are released during radioactive decay? alpha particles be

USPshnik [31]

The answer is B) beta particles

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3 years ago

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