All categories

3 years ago

Which question could be tested in a scientific manner?

1 answer:

andrey2020 [161]3 years ago

5 0

The questions from this lot which could be tested in a scientific manner would be "what causes some people to be color-blind"
and
"what are the best shoes to wear when exercising"
Both of these questions can be tested in a scientific way through an experiment.

You might be interested in

What is the speed of a bobsled whose distance-time graph indicates that it traveled 113m in 29s?

Elza [17]

In this item, we are asked to determine the speed of the bobsled given the distance traveled and the time it takes to cover the certain distance. This can mathematically be expressed as,
speed = distance / time

Substituting the given values in this item,
speed = (113 m) / (29 s)
speed = 3.90 m/s

ANSWER: 3.90 m/s

3 0

3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago

A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhauste

marusya05 [52]

516.154 megawatts of heat are exhausted to the river that cools the plant.

By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river ( $Q_{out}$ ), in megawatts: