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4 years ago

6

Two groups of students were tested to compare their speed working math problems,Each group was given the same problems.One group

used calculators and the other group computed without calculators .what is the hypothesis?

1 answer:

Zanzabum4 years ago

4 0

Your hypothesis can be that the group with calculators would finish faster than the group without calculators.

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3 years ago

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In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.

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3 years ago

Changes in behavior of managers, coworkers, and subordinates can be a warning sign before a layoff. True or false

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The correct answer is - true.

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3 years ago

What is a force that resist the motion of one force over another

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3 years ago

Read 2 more answers

A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a

photoshop1234 [79]

Answer:

a) f' = 432 Hz

b) I = 8.12*10^-4 W/m^2

Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

$f'=f(\frac{v-v_o}{v+v_s})$ (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

$v=\lambda f\\\\$

v: speed of sound

λ: wavelength = 0.700 m

$f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz$

Next, you replace the values of all parameters in the equation (1):

$f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz$

hence, the frequency perceived by the car is 432 Hz

b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:

$I=\frac{P}{4\pi r^2}$

P: power of the source = 50.0 W

r: distance to the source = 70.0 m

$I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}$

hence, the intensity is 8.12*10^⁻4 W/m^2

3 0

3 years ago