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frutty [35]
3 years ago
10

A researcher studying the nutritional value of a new candy places a 4.60 g sample of the candy inside a bomb calorimeter and com

busts it in excess oxygen. The observed temperature increase is 2.69 ∘ C. If the heat capacity of the calorimeter is 32.20 kJ ⋅ K − 1 , how many nutritional Calories are there per gram of the candy?
Physics
1 answer:
blagie [28]3 years ago
8 0

Answer:

4500.5 nutritional calories per gram

Explanation:

Heat lost by the new candy = heat gained by the bomb calorimeter.

Heat gained by the bomb calorimeter = c×ΔT

where c = heat capacity of the calorimeter = 32.20 KJ/K = 32200 J/K

ΔT = change in temperature = 2.69°C = 2.69 K.

Heat gained by the bomb calorimeter = 32200 × 2.69 = 86618 J

Heat lost by the new candy = heat gained by the bomb calorimeter = 86618 J = 20702.2 calories

4.60 g of the new candy lost this amount of calories by undergoing combustion,

The amount of calories per g = 20702.2 calories/4.6 g = 4500.5 calories per gram

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Explanation:

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3 years ago
With the aid of a suitable diagram, explain what is meant by an electromagnetic wave.
suter [353]

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Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields.

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3 years ago
Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo
kifflom [539]

Answer:

893 Newtons

Explanation:

m = Mass of Usain Bolt = 94 kg

a = Acceleration of Usain Bolt in the first 0.89 seconds = 9.5 m/s²

From Newton's Second law

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F=ma

\\\Rightarrow F=94\times 9.5

\\\Rightarrow F=893\ N

The average horizontal force exerted by Bolt against the ground during the first 0.890 s of the race is 893 Newtons

3 0
3 years ago
An image of a car with height of 14 cm occurred in the mirror which is located at a T-
Thepotemich [5.8K]

Answer:

The distance of car form the mirror is 330 cm.

Explanation:

height of object, h = 140 cm

height of image, h' = 14 cm

radius of curvature, R = 60 cm

focal length, f = R/2 = + 30 cm

Let the distance of image is v and the distance of object is u.

\frac{h'}{h}\frac{v}{u}\\\\\frac{14}{140} =\frac{v}{u}\\\\v =\frac{u}{10}

Use the formula of focal length

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{30}=\frac{10}{u}+\frac{1}{u}\\\\\frac{1}{30}=\frac{11}{u}\\\\u = 330 cm

3 0
3 years ago
Give an example of a measurement that is precise to the nearest tenth of a gram.
Svet_ta [14]

Answer:

So we have a measure in grams, we can start with something like:

143.523 grams.

Now we want this measurement to be precise to the nearest tenth of a gram.

The nearest tenth of a gram is the first digit after the decimal point, then the digits that come after this are not useful, because they are outside our precision range.

Then we must write our measurement as:

143.5 grams

Where the digit that came after the 5 was a 2, so we rounded down.

4 0
3 years ago
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