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Irina-Kira [14]
3 years ago
10

How do you know if an experimental result is acceptable or trustworthy? What gives you confidence that your data are trustworthy

?
Physics
1 answer:
Virty [35]3 years ago
6 0

Explanation:

For an experimental result to be considered acceptable, all relevant variables involved in the experiment must be taken into account, by isolating it, performing it under controlled conditions and modifying the conditions under which it takes place. This, with the objective of excluding alternative explanations in the analisis of the experimental data. Therefore, if these steps are followed appropriately, experimental data are trustworthy. The reliability of the experiment increases when it is replicated by other researchers and the same results are obtained.

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How long does it take the projectile to reach its highest point in its trajectory?​
NeTakaya

Answer:

It takes about 88 seconds for the cannonball to reach its maximum height (ignoring air resistance).

Explanation:

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3 years ago
Simple diffusion and facilitated diffusion are related in that both
kaheart [24]

Answer:

allow the downward movement of the concentration gradient by passive transport

Explanation:

Passive transport is a process of substance transport, which is carried out spontaneously, without energy expenditure and in favor of the concentration gradient, that is, from a medium where the molecules are more concentrated towards a medium where their concentration is lower.

Three types of passive transport are distinguished: osmosis, simple diffusion and facilitated diffusion

<u>Simple diffusion</u>

It is the passage, through the plasma membrane, of small molecules without charge soluble in the lipid bilayer, such as some gases (oxygen and carbon dioxide). For a molecule to diffuse through the membrane it is necessary that there is a difference in concentration between the external and the internal environment.

<u>Diffusion facilitated </u>

There are molecules such as amino acids, glucose and small ions that, due to their chemical and size characteristics, cannot diffuse through the lipid bilayer and require transport proteins for diffusion.

The transport proteins are immersed in the plasma membrane and can be of two types: protein channels, formed by proteins that generate a channel in the membrane, and permeases, which are proteins that, when joined to the molecule to be transported, change their shape by carrying them into the cell.

4 0
3 years ago
Consider an electromagnetic wave of frequency f=3x10^6 Hz. does this radiation belong to the visible range to the ultraviolet or
In-s [12.5K]

Answer:

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Explanation:

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3 0
3 years ago
How does velocity differ from speed? What changes in motion can result in a change in velocity?
olga55 [171]
-Velocity is the speed of any moving object in a given direction, whilst Speed is the rate of an object's ability to move.
-Velocity can change if the direction or time is changed, the basic equation of velocity is: v = d/t
v - velocity
d - distance
t - time
If one of these factors change, it affects the other.

Hope this answers the question!
4 0
3 years ago
Read 2 more answers
A car's bumper is designed to withstand a 7.20 km/h (2.0 m/s) collision with an immovable object without damage to the body of t
ollegr [7]

Answer:

8512 N

Explanation:

From the work energy theorem we know that: The net work done on a particle equals the change in the particles kinetic energy:

W=\Delta K=K_{f}-K_{i} \\&#10;\\qquad \begin{array}{r}&#10;W=F \cdot d, \Delta K=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\&#10;F \cdot d=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}

Where:

-W is the work done by the force.

- F is the force actin on the.

- d is the distance travelled.

- m is the mass of the car.

- v_{f}, v_{i} are the final and the initial velocity of the car

K_{f}, K_{i} are the final and the kinetic energy of the car.

Givens: m=830 \mathrm{~kg}, v_{i}=2 \mathrm{~m} / \mathrm{s}, v_{f}=0 \mathrm{~km} / \mathrm{h}, d=0.195 \mathrm{~m}

Plugging known information to get:

F \cdot d &=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2} \\&#10;F &=\frac{\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}}{d} \\&#10;&=\frac{0-\frac{1}{2} \times 830 \times 2^{2}}{0.195} \\&#10;&=8.512 \times 10^{3} \\&#10;F &=8.512 \times 10^{3} \mathrm{~N}

8 0
3 years ago
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