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Gnoma [55]
3 years ago
12

Buttery popcorn contained in a large 1__bowl has a mass of about 50 __and about 650 calories.​

Physics
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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A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108
tiny-mole [99]

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

W = F*d

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

W = F*d = 108.915 N*2.38 m = 259.22 J

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

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3 years ago
PLS HELP :) GIVING BRAINLIEST SIMPLE SCIENCE QUESTION HELPS PLSSSS
raketka [301]
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3 years ago
Nathalie leaves a history classroom and walks 3 meters North to drinking fountain. Then she turns and walks 10 meters south to a
pishuonlain [190]

Answer:

13 meters

Explanation:

Step one:

given

We are told that Nathalie leaves a history classroom and walks 3 meters North

Then travels another 10 meters south to an art classroom.

Required

The total distance.

Step two:

The total distance can be computed by summing up the 3 meter North distance traveled and the 10 meter south distance traveled

Total distance= 3+10= 13meters

7 0
3 years ago
A flying saucer lifts the Physical Science building 10,000 ft into the air before discovering it is useless and discards the rem
scZoUnD [109]

Answer:

500000000 lbft/s

Explanation:

F = Force or weight = 1000000 lbf

s = Displacement = 10000 ft

t = Time taken = 20 seconds

Work done is given by

W=Fs\\\Rightarrow W=1000000\times 10000\\\Rightarrow W=10000000000\ lb-ft

Power is given by

P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{10000000000}{20}\\\Rightarrow P=500000000\ lbft/s

One Saucer power is 500000000 lbft/s

7 0
3 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
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