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3 years ago

If a calorie is equivalent to 4.184 joules how many joules are contained in 250 kg calories slice of pizza

1 answer:

6 0

In order to answer this, we will set up a simple ratio as such:

1 calorie = 4.184 joules

1 kilocalorie = 1000 calories

1 kilocalorie = 4,184 joules

250 kilocalories = x joules

Cross multiplying the second and third equations, we get:

x joules = 4,184 * 250

250 kilocalories are equivalent to 1,046 kJ

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a 202 kg bumper car moving right at 8.50 m/s collides with a 355 kg car at rest. Find the total momentum of the system.

Hunter-Best [27]

Explanation:It is given that,

Mass of bumper car, m₁ = 202 kg

Initial speed of the bumper car, u₁ = 8.5 m/s

Mass of the other car, m₂ = 355 kg

Initial velocity of the other car is 0 as it at rest, u₂ = 0

Final velocity of the other car after collision, v₂ = 5.8 m/s

Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁

Using the conservation of linear momentum as :

p₁ = m₁v₁ = -342 kg-m/s

So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.

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3 years ago

Read 2 more answers

Lightning and thunder take place in the sky at the same time and at the same distance from us. Lightning is seen earlier and thu

Tasya [4]

Answer:

Lightning is seen earlier than the thunder because speed of light is more than the speed of sound. Therefore, even though both occurs as same time and place in the sky, lightning is seen earlier.

3 0

3 years ago

A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng

frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

$Q_{1} =\frac{W}{n}$

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

$Q_{1} =\frac{750}{0.35} =2142.86kJ$

$Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ$

c) The efficiency of Carnot engine is:

$n=1-\frac{300K}{550K} =0.455$

The heat transfer to the heat engine is:

$Q_{1c} =\frac{750}{0.455} =1648.35kJ$

The heat transfer from the heat engine is:

$Q_{2c} =1648.35-750=898.35kJ$

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3 years ago

Where does the basketball have the greatest gravitational potential energy?

Olenka [21]

The basketball has the greatest gravitational potential energy at its, highest which is at the very top.

6 0

3 years ago

A cyclist makes the following trip along two vectors; he travels 9km to the north and then travels 6km to the east

elena-14-01-66 [18.8K]

Answer:

Final distance from the origin: 10.82 km. the vector points as shown in the attached image.

Angle with respect to the east: $56.31^o$

Explanation:

Please refer to the attached image. The cyclist's trip is indicated with the green arrows (9 km to the north followed by 6 km to the east.

So his final position is at the tip of this last vector, and indicated by the orange vector drawn form the point where the trip starts to the cyclist's final location.

We observe that this orange vector is in fact the hypotenuse of a right angle triangle, and we can estimate the distance from the origin by the Pythagorean theorem:

$d=\sqrt{9^2+6^2} \\d=\sqrt{81+36} \\d=\sqrt{117} \\d=10.82 \,\,km$

Notice that this is NOT the actual number of km that the cyclist pedaled to reach the final point.

Now, to find the value of the angle $\theta$ , we need to use trigonometry, and in particular the tangent function gives us the ratio between the side of the triangle "opposite" to the angle, divided the side "adjacent" to the angle:

$tan(\theta)=\frac{opp}{adj} \\tan(\theta)=\frac{9}{6}\\tan(\theta)=\frac{3}{2}\\$

Now we can find the value of the angle by using the arctan function:

$tan(\theta)=\frac{3}{2} \\\theta=arctan(\frac{3}{2} )\\\theta= 56.31^o$

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3 years ago