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3 years ago

An insulated lunch bag keeps food warm by

Physics

2 answers:

Greeley [361]3 years ago

8 0

Answer: slowing the transfer of thermal energy from the food inside the bag to the air outside the bag.

Hatshy [7]3 years ago

4 0

I believe the answer is D

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Laney et al. (false memory):(a) Outline the rating scale used in the Food History Inventory [2 marks]

djverab [1.8K]

Answer: The rating scale that is used in food history inventory is from 1 to 8.

Explanation:

In this case the participants have to rate 24 items on a scale of 1 to 8.There are variety of questions related to food or fooding habits.

The questions like do you go out for food?

Do you like bread ? if yes how will you rate it on a scale of 1 to 8.

The questions like this has to be answered in a way . It is done in order to see the preferences and false memory regarding the food preference that people have.

6 0

3 years ago

Which subatomic particle can be absent from an atom?

Alika [10]

Neutron can be absent

5 0

3 years ago

The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k

Brrunno [24]

Answer:

The gravitational force changing velocity is

$\frac{dF}{dt}=-8\frac{N}{s}$

Explanation:

The expression for the gravitational force is

$F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}$

Differentiate the above equation

$\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}$

The velocity is the distance in at time so

$V=\frac{dr}{dt}=0.4 \frac{km}{s}$

$\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}$

$\frac{dF}{dt}=-8\frac{N}{s}$

8 0

3 years ago

69PTS! Which choices are also major levels of classification? Choose all answers that are correct. A. Genus B. Species C. Group

Pachacha [2.7K]

A. Genus and B. Species are also major levels of classification.

5 0

3 years ago

Read 2 more answers

Alan leaves Los Angeles at 8:00 A.M. to drive to San Francisco 400 mi away. He manages to travel at a steady 50 mph in spite of

Fudgin [204]

Answer:

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

Explanation:

let 'd' be distance b/w Los Angeles and San Francisco i.e 400 mi

considering ,

Alan's speed $v_A$ =50mph

Beth's speed $v_B$ =60mph

->For Alan:

The time required $t_A$ = d/ $v_A$ = 400/50 => 8h

-> For beth:

The time required $t_B=\frac{d}{v_B} =\frac{400}{60} =>6\frac{2}{3} h$ => 6h 40m

Alan will reach at 8:00 a.m +8h = 4:00p.m.

Beth will reach at 9:00 a.m +6h 40m= 3:40p.m.

a) Beth will reach before Alan

b)Beth has to wait 20 min for Alan to arrive

3 0

3 years ago