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aleksandrvk [35]
3 years ago
12

A researcher studying the nutritional value of a new candy places a 6.60 g 6.60 g sample of the candy inside a bomb calorimeter

and combusts it in excess oxygen. The observed temperature increase is 2.46 ∘ C. 2.46 ∘C. If the heat capacity of the calorimeter is 33.90 kJ ⋅ K − 1 , 33.90 kJ⋅K−1, how many nutritional Calories are there per gram of the candy?
Physics
1 answer:
cricket20 [7]3 years ago
7 0

Answer:

there are 3.018 kcal= 3018 cal per gram of candy

Explanation:

If the assume that the calorimeter is perfectly insulated, then all the heat released by the combustion is absorbed by the calorimeter.

Also knowing that Q= C * ΔT , where C= heat capacity of the calorimeter , ΔT= temperature change , Q = heat released by the combustion of the candy

replacing values

Q = C * ΔT = 33.90 kJ/°C * 2.46°C = 83.394 kJ

since Q is the heat released when burned all the mass m of the candy, the number of calories per gram of candy will be

q = Q/m =83.394 kJ / 6.60 g = 12.635 kJ/g

q = 12.635 kJ/g * 1 kcal / 4.186 kJ = 3.018 kcal per gram of candy

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Answer:

2.9 m

Explanation:

First find the time it takes to reach the floor.

y = y₀ + v₀ t + ½ at²

(0 m) = (1.6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 0.571 s

Next, find the distance it travels in that time.

x = x₀ + v₀ t + ½ at²

x = (0 m) + (5.0 m/s) (0.571 s) + ½ (0 m/s²) (0.571 s)²

x = 2.86 m

Rounded to two significant figures, the marble travels 2.9 meters in the x direction.

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Which is a device that stores electric charge by separating positive and negative charges?
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A young boy is dragging his 10 kg stuffed animal by the tail at a constant velocity of 4 m/s. The coefficient of friction betwee
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3 years ago
1. An object on Earth and the same object on the Moon would have a difference in
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Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.

<h2>Answer 1: a. weight</h2>

Mass and weight are very different concepts.  

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.  

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass m of the body by the acceleration of gravity g:  

W=m.g  

Then, since the Earth and the Moon have different values ​​of gravity, t<u>he weight of an object in each place will vary</u>, but its mass will not.

<h2>Answer 2: b. Force changes by 2/9</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:  

F=G\frac{m_{1}m_{2}}{r^2} (1)

Where:  

F is the module of the force exerted between both bodies  

G is the universal gravitation constant

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

If we double the mass of one object (for example 2m_{1}) and triple the distance between both (for example 3r). The equation (1) will be rewritten as:

F=G\frac{2m_{1}m_{2}}{(3r)^2} (2)

F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2} (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

<h2>Answer 3: b. movement</h2>

The Work W done by a Force F refers to the release of potential energy from a body that is <u>moved</u> by the application of that force to overcome a resistance along a path.  

When the applied force is constant and <u>the direction of the force and the direction of the movement are parallel,</u> the equation to calculate it is:  

W=(F)(d)

Now, <u>when they are not parallel, both directions form an angle</u>, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha}

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

<h2>Answer 4: a. 40,000 Joules</h2>

The Kinetic Energy is given by:

K=\frac{1}{2}mV^{2}   (4)

Where m is the mass of the body and V its velocity

For the first case (kinetic energy K_{1}=10000J  for a car at V_{1}=30 mph=13.4112m/s):

K_{1}=\frac{1}{2}mV_{1}^{2}   (5)

Finding m:

m=\frac{2K_{1}}{V_{1}^{2}}   (6)

m=\frac{2(10000J)}{(13.4112m/s)^{2}}   (7)

m=111.197kg   (8)

For the second case (unknown kinetic energy K_{2}  for a car with the same mass at V_{2}=60 mph=26.8224m/s):

K_{2}=\frac{1}{2}mV_{2}^{2}   (9)

K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}   (10)

K_{2}=40000J   (11)

<h2>Answer 5: c. the soil will be 5°C</h2>

The formula to calculate the amount of calories Q is:

Q=m. c. \Delta T   (12)

Where:

m  is the mass

c  is the specific heat of the element. For water is c_{w}=1 kcal/g\°C  and for soil is c_{s}=0.20 kcal/g\°C  

\Delta T  is the variation in temperature (the amount we want to find for both elements)

This means we have to clear \Delta T from (12) :

\Delta T=\frac{Q}{m.c}   (13)

For Water:

\Delta T_{w}=\frac{Q_{w}}{m_{w}.c_{w}}   (14)

\Delta T_{w}=\frac{1kcal}{(1kg)(1 kcal/g\°C)}   (15)

\Delta T_{w}=1\°C)}   (16)

For Soil:

\Delta T_{s}=\frac{Q_{s}}{m_{s.c_{s}}   (17)

\Delta T_{s}=\frac{1kcal}{(1kg)(0.20 kcal/g\°C)}   (18)

\Delta T_{s}=5\°C)}   (19)

Hence the correct option is c.

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Answer:

14.035087719298246 ≈ 14 hours

Explanation:

just subtract the 4000 and the 11000 from 23000 which gives you 8000. And then divide that my 570

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