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3 years ago

What is the difference between the control group and the experimental group in an experimental study?

1 answer:

6 0

Answer:a control group has no effects or changes whilst the experimental group has changes or something different from a control group

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Science question please help me

S_A_V [24]

Light year is light in a year so 4 light years are 4 Earth years

5 0

3 years ago

Temperatures expressed in the Kelvin scale are ____ higher than temperatures in the Celsius scale?

Cerrena [4.2K]

Answer: D. 273

Explanation: Celsius temperatures can be negative while kelvin goes down to absolute zero and doesn’t go any lower, thus no negative numbers

4 0

3 years ago

How much net force is required to accelerate a 5 kg toy car, initially at rest to a velocity of 3 m/s in 6 s?

inna [77]

Answer:

force = 0.20N .F = m ×a .& a = v/t then the f = m×v/t.

Explanation:

4 0

3 years ago

A camera phone called the iPhone X has an image sensor size of 5 mm and a focal length of 4 mm. How long of a selfie-stick must

azamat

Answer:

length of selfie-stick is 1.62 m

Explanation:

Given data

image size h1 = 5 mm = 5 × $10^{-3}$ m

focal length = 4 mm = 4 × $10^{-3}$ m

distance h2 = 2.032 m

to find out

How long of a selfie-stick

solution

here we find first magnification

that is M = h1 /h2

M = 5 × $10^{-3}$ / 2.032

M = 2.46 × $10^{-3}$

and we know M = p/q

so p = Mq = 2.46 × $10^{-3}$ q

so we apply lens formula

1/f = 1/p - 1/q

1/ 4 × $10^{-3}$ = 1 / 2.46 × $10^{-3}$ q - 1/q

q = 1.622 m

so length of selfie-stick is 1.62 m

4 0

3 years ago

Frank’s automobile engine runs at 100∘C. One day, when the outside temperature is 15∘C, he turns off the ignition and notes that

lapo4ka [179]

Answer:

the engine cool to 40 $^{o}C$ at 14.07 minutes

Explanation:

Given information

T(5) = 70 $^{o}C$

$T_{0}$ = 100 $^{o}C$

C = 15 $^{o}C$

Newton's law of cooling :

T(t) = C + ( $T_{0}$ - C) $e^{-kt}$

where

T(t) = temperature at any given time

C = surrounding temperature

$T_{0}$ = initial temperature of heated object

k = cooling constant

to find the the time when the engine will be cooled down to 40 $^{o}C$ , we first need to find the cooling constant, k

when t = 5, T(5) = 70 $^{o}C$

so,

T(t) = C + ( $T_{0}$ - C) $e^{-kt}$

T(5) = 15 + (100 - 15) $e^{-5k}$

70 = 15 + (85) $e^{-5k}$

$e^{-5k}$ = (70 - 15) / 85

-5k = ln (55/85)

k = - ln (55/85) / 5

k = 0.087

thus, we have the eqaution

T(t) = 15 + (85) $e^{-0.087t}$

now we can determine the time when T(t) = 40 $^{o}C$

40 = 15 + (85) $e^{-0.087t}$

$e^{-0.087t}$ = (40-15)/85

-0.0087t = ln (25/85)

t = - ln (25/85)/0.087

t = 14.07 minutes

8 0

3 years ago