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Elena L [17]
3 years ago
15

Which of the following is true about organizing items and information during scientific investigations?

Physics
2 answers:
LekaFEV [45]3 years ago
5 0
There are usually multiple methods for organizing items and information in a scientific investigation.
frez [133]3 years ago
3 0

Answer:

There are usually multiple methods for organizing items and information in a scientific investigation.

Explanation:

<em>There are usually multiple methods for organizing items and information in a scientific investigation.</em> For example, a scientist might keep separate records for each trial he or she runs in an experiment. Alternatively, records for all trials could be kept on one data sheet. Records might be organized by date. In other cases, they might be organized based on patterns in experimental results.

Scientists also use multiple methods for displaying and analyzing data. For example, collected data in one investigation might best be displayed in a data table. In other cases, plotting data on a graph or chart might help the scientists better understand the data.

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A football player kicks a 0.94 kg football with a force of 2.4 N. Calculate the acceleration of the football as the player kicks
Anon25 [30]
Hello, I hope this helps :)
So the equation to figure these kinds of questions is F=MA
F refers to force, which in this situation F would be 2.4
M refers to mass, mass would be 0.94
A refers to acceleration, which we are trying to figure out
So if we put in the information we know into the equation, it is now 2.4=0.94*A
So with that equation we can figure that 2.4 divided by 0.94 equals A
So 2.4/0.94= 2.553191489362 :'D
Don't worry, the rounded and correct answer is 2.6
Have a nice day =)
7 0
2 years ago
Read 2 more answers
What does this circuit do? T is the clock All flip flops are D-Flip-Flop What is the function of the circuit if Dser is ‘1’ alwa
Harlamova29_29 [7]

Answer:

That is true

Explanation:

6 0
2 years ago
A wheelbarrow is a good example of a second-class lever. True or False
olganol [36]

Answer:

true

Explanation:

a wheelbarrow has its load situated between the fulcrum and the force the wheel Barrow is 2nd class because of its resistance between the force and the axis

3 0
3 years ago
The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ
Marizza181 [45]

The energy stored in the membrane is 6.44\cdot 10^{-14} J

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

where

k is the dielectric constant of the material

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

A=4.50\cdot 10^{-9} m^2

d=8.1\cdot 10^{-9} m

Substituting, we find its capacitance:

C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F

Now we can find the energy stored: for a capacitor, it is given by

U=\frac{1}{2}CV^2

where

C=2.26\cdot 10^{-11} F is the capacitance

V=7.55\cdot 10^{-2} V is the potential difference

Substituting,

U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

6 0
3 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
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