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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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Which of the following is a component of staying positive while competing ​
Natalija [7]

Answer: B - complimenting others on good plays

Explanation: Reading the first words sort of give it away when staying positive you compliment, not criticize, confront angrily, or refuse.

4 0
3 years ago
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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
2 years ago
Answer this question ASAP please and thank you
allochka39001 [22]
The answer is Oscar Robertson and Jerry West !! hope this helps :p
8 0
3 years ago
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When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play
saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

6 0
3 years ago
Which statement accurately describes a sample of water during parts a and c of the heating curve
vivado [14]

Answer:

A and C is about 12 cm away from each other.

Explanation:

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3 years ago
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