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Kisachek [45]
1 year ago
13

Anita is comparing the accepted value for a physical property to the value she measured in the laboratory. Which characteristic

of her measured value is Anita most likely trying to evaluate? accuracy mean precision range
Physics
2 answers:
Butoxors [25]1 year ago
7 0

Answer:

nita must look for the percentage error and look for the precision of its measurement.

Explanation:

When a measurement of a physical property is made in the laboratory, it must be compared with the accepted values. For this, the mean value of the laboratory measurement is calculated and the absolute or percentage error is sought with respect to the accepted value.

In conclusion Anita must look for the percentage error and look for the precision of its measurement.

sp2606 [1]1 year ago
5 0

Answer:

accuracy

Explanation:

correct on edge2020

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chubhunter [2.5K]

Explanation:

Whole system will accelerate under the action of applied force. The box will experience the force against the friction and when this force exceeds then the box will move. so

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The applied force is given by

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2 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
2 years ago
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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

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To find,

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Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

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k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

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