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3 years ago

All of the following activities produce quantitative data exept

Physics

1 answer:

Keith_Richards [23]3 years ago

7 0

Answer:

The answer to your question are A and C

Explanation:

Quantitative data are quantities, something that we get after measuring something.

A. Measuring the rate of gas production from a chemical. This example is a quantitative measure, because we are measuring the rate.

B. Describing the clarity of water in a sample If we are describing something, means that we are not measuring anything, so this is not a quantitative measure.

C. Calculating the energy released from an electrochemical reaction If we are not measuring but we are using the data somebody else got to calculate energy, them this is a quantitative data.

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It took 450/340 = 1.32 seconds

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17. What is the speed of a race horse that runs<br> 1500 m in 125 s?

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12 miles per second

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3 years ago

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Problem 5 You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this gam

Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

The object distance is $y= 6in$ .

Because the surface is flat, the radius of curvature is infinity .

The incident index is $n_i=\frac{4}{3}$ and the transmitted index is $n_t= 1$ .

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Substituting the quantities given in the problem,

$\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0$

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3 years ago

A basketball is tossed upwards with a speed of 5.0 m/s. We can ignore air resistance.

FinnZ [79.3K]

We have that the maximum height reached by the basketball from its release point is

$S=1.3m$

From the question we are told

A basketball is tossed upwards with a speed of 5.0 m/s. We can ignore air resistance.
What is the maximum height reached by the basketball from its release point?

Generally the Newtons equation for Motion is mathematically given as

$V^2=u^2+2as\\\\Therefore\\\\0=25-19.6*S\\\\S=\frac{25}{19.6}\\\\S=1.276$

$S=1.3m$

Therefore

The maximum height reached by the basketball from its release point is

$S=1.3m$

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3 years ago

Two musicians are comparing their trombones. The first produces a tone that is known to be 438 Hz. When the two trombones play t

galben [10]

Answer:

It is producing either a 435-Hz sound or a 441-Hz sound.

Explanation:

When two sound of slightly different frequencies interfere constructively with each other, the resultant wave has a frequency (called beat frequency) which is equal to the absolute value of the difference between the individual frequencies:

$f_B = |f_1 -f_2|$ (1)

In this problem, we know that:

- The frequency of the first trombone is $f_1 = 438 Hz$

- 6 beats are heard every 2 seconds, so the beat frequency is

$f_B=\frac{6}{2 s}=3 Hz$

If we insert this data into eq.(1), we have two possible solutions for the frequency of the second trombone:

$f_2 = f_1 - f_B = 438 Hz - 3 Hz = 435 Hz\\f_2 = f_1+f_B = 438 Hz+3 Hz=441 Hz$

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3 years ago