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3 years ago

12

Buttery popcorn contained in a large 1bowl has a mass of about 50 and about 650 calories.

2 answers:

Fudgin [204]3 years ago

6 0

Answer:

Buttery popcorn contained in a large 1 <u>liter</u> bowl has a mass of about 50 <u>grams</u> of fat and about 650 calories

Explanation:

The question is about using correct units. The capacity of a bowl must be measured in a volume unit; the options were milliliter and liter. 1 milliliter is a very small amount (about 0.002 pint), so 1 liter is the correct option.

For the second blank, the options (mass units) are: microgram, milligram, gram, kilogram. Milligram is used to measure the mass of minerals, microgram is even smaller than milligram, the buttered popcorn weights much less than 50 kilograms. So, the correct option is 50 grams.

Rasek [7]3 years ago

4 0

Answer:

Buttery popcorn contained in a large 1 liter bowl has a mass of about 50 mg and about 650 calories.

Explanation:

Liter is the most appropriate unit to measure a bowl. Usually 1 liter of liquid has a mass of 1000 gram. Since popcorn is something lightweight and only a few can fill the bowl quickly so 50 mg makes perfect sense with 1 liter of bowl and 650 calories in buttery popcorn.

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Two samples of water are mixed together.

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Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

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Givens

Ice

deltat = x

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deltat = 100 - x

m = 1.5 kg

c = 4.18

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The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x) Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x) Remove the brackets.

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Answer

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3 years ago

A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple

aliina [53]

Answer:

T= 1 s

Explanation:

Given that

When x= cm ,T= 1

we know that time period of spring mas system given as

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T= Time period

m= mass

k=spring constant

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Explanation:

It is given that,

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$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

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