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Oliga [24]
2 years ago
6

Give two examples of when you would need to use the measure skill in science?

Physics
1 answer:
Iteru [2.4K]2 years ago
6 0
I'm guessing "measure skill" means "the ability to measure things." In reality, all experiments by necessity require data and typically we need to measure things to get them (even if this is done by devices, programs or computers). When doing science labs, you'll likely need to use scales, pipets, and various glasswear to measure different things. Even if you're used to using a ruler, getting a really good measurement that you can feed into equations and get meaningful results from requires a bit of practice and more care than you might think. I'd also say that the measurement skill comes into play when making approximations or assumptions about experiment. No measurement is infinitely accurate, you can't measure the width of an atom with a standard 12 inch ruler, or if you did, you'd have to have a very large amount of error. Making these logical conclusions about your devices, where they reach their limits, and what potential error you may have and where it comes from are all important when doing science.
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A ball is thrown downward with an initial speed of 7m/s. the balls velocity after 3 seconds is m/s
ycow [4]

Answer:

velocity = distance / time

distance = speed * time

Distance= 7*3 = 21

V = 21/3

V = 7

Explanation:

4 0
2 years ago
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A thin conducting square plate 1.0 m on the side is given a charge of-2.0 x 10-6 c. A proton is placed 1.0 en above the center o
adell [148]

Answer:

Acceleration, a=1.08\times 10^{13}\ m/s^2

Explanation:

It is given that,

Side of the square plate, l = 1 m

Charge on the square plate, Q=-2\times 10^{-6}\ C

Position of a proton, x = 1 cm

The electric field due to a parallel plate is given by :

E=\dfrac{Q}{2A\epsilon_o}

Electric force is given by :

F = q E

F=\dfrac{Qe}{2A\epsilon_o}

e is the charge on electron

The acceleration of the proton can be calculated as :

a=\dfrac{F}{m}

m is the mass of proton

a=\dfrac{Qe}{2A\epsilon_o m}

a=\dfrac{2\times 10^{-6}\times 1.6\times 10^{-19}}{2(1)^2\times 8.85\times 10^{-12}\times {1.67\times 10^{-27}}}

a=1.08\times 10^{13}\ m/s^2

So, the acceleration of the proton is 1.08\times 10^{13}\ m/s^2. Hence, this is the required solution.

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3 years ago
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ololo11 [35]

I think the answer is C

Hope this help

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Answer:

your

answer is

12.004

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