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3 years ago

A problem solving method that involves trying all possible solutions until one works is using_____.

Physics

1 answer:

skad [1K]3 years ago

6 0

Answer:

trial and error.

Explanation:

a problem solving method that involves trying all possible solutions until one works is using trail and error.

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Ions can cross a membrane without the help of an integral membrane protein. True or False

denis-greek [22]

Answer:

False

Explanation:

ionic particles, which are hydrophilic, cannot easily cross a membrane. Very small polar molecules, such as water, can cross via simple diffusion due to their small size. Charged atoms or molecules of any size cannot cross the cell membrane via simple diffusion as the charges are repelled. Because these ions can cross the cell membrane, their movement is restricted to protein channels and specialized transport mechanisms in the membrane.

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4 years ago

Which of the following does not accurately describe the forces that exist with an atom? A. Electrons in an atom are attracted to

LenKa [72]

<span>A, B, and C are correct descriptions.

Choice-D does NOT accurately describe the forces
that exist within an atom.

Choice-D should say:
"Electrons positioned closer to the nucleus have a greater
attraction to the protons and are LESS likely to be discharged
from the atom than electrons farther away are."</span>

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3 years ago

Its ____ makes Sirius the brightest star in the night sky.

grigory [225]

With its apparent magnitude

7 0

3 years ago

Read 2 more answers

The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

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3 years ago

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X-ray)
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3 years ago