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4 years ago

13

The _____ is the process scientists use to conduct research, which includes a continuing cycle of exploration, critical thinking

, and systematic observation.
experimental method

scientific method

subjective experience

Institutional Review Board

2 answers:

bonufazy [111]4 years ago

8 0

I believe it’s the scientific method or either the experimental method

trasher [3.6K]4 years ago

6 0

I believe it is the scientific method

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The teachers of Scott, Christopher, Dianne, and Kailee have last names Koeninger, Dannemiller, Briscoe, and Carter. Match the ki

Pani-rosa [81]

Scott-Dannemiller, Koeninger, Briscoe, and Carter

Christopher-Briscoe, Dannemiller, and Koeninger

Dianne-Koeninger, Briscoe, and Carter

Kailee-Koeninger and Carter

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4 years ago

A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it ha

dlinn [17]

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

$a=(-2v^3)\ m/s^2$

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=-2v^3$

$\dfrac{dv}{-v^3}=2dt$

On integrating

$int{\dfrac{dv}{-v^3}}=\int{2dt}$

$\dfrac{1}{2v^2}=2t+C$

$v^2=\dfrac{1}{4t+2C}$ ....(I)

At t = 0, v = 10 m/s

$10^2=\dfrac{1}{4\times0+2C}$

$C=\dfrac{1}{200}$

Put the value of C in equation (I)

$v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}$

$v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}$

$v=0.099\ m/s$

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

$\dfrac{ds}{dt}=v$

On integrating

$\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt$

$s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'$

At t = 0, s = 15 m

$15=\dfrac{200}{800}+C'$

$C'=15-\dfrac{200}{800}$

$C'=14.75$

Put the value in the equation

$s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75$

$s=19.75\ m$

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

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3 years ago

List the three classes of rock and name a specific rock that goes with that class

Vsevolod [243]

Metamorphosis that's all I know

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4 years ago

What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate

Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The values is $E =248.2 \ N/C$

Explanation:

From the question we are told that

The diameter is $d = 12 \ cm = 0.12 \ m$

The charge is $Q = 4.4 nC = 4.4 *10^{-9} \ C$

The distance from the center is $k = 0.1 \ mm = 1*10^{-4} \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.12}{2}$

=> $r = 0.06 \ m$

Generally electric field is mathematically represented as

$E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]$

substituting values

$E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]$

$E =248.2 \ N/C$

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4 years ago

11. How many marine mammals from Figure 1 can hear sounds under 100 Hz? 12. If there was a sound that was 100,000 Hz, which mamm

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Explanation:

i REALLY HOPE you get this DUB

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3 years ago