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3 years ago

11

The fact that our preconceived ideas contribute to our ability to process new information best illustrates the importance of: th

e serial position effect. O repression iconic memory . semantic encoding . retroactive interference .

1 answer:

lyudmila [28]3 years ago

3 0

Answer:

It’s a

Explanation:

Don’t actually put that i needed the points mb

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3 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

NEED HELP ASAP!!!!!!!!!!!!!

love history [14]

Answer: Decreased risk of heart attack

Explanation:

thats the answer because it actually

increase rish of heart attack

Hope this helps :)

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4 years ago

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