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kolbaska11 [484]
2 years ago
8

A piece of candy has 5 calories (or 5000 calories). if it could be burned, leaving nothing but carbon dioxide and water, how muc

h heat would it give off?
Physics
2 answers:
posledela2 years ago
8 0
A) 500 calories 
B)5000 joules
C) 5 kilocalories 

Answer is C
satela [25.4K]2 years ago
5 0

Answer:

5 kilocalories

Explanation:

The heat is a way of energy and it can be measured by calories, but the calories of the heat (thermal calories) are different for the food Calories.

1 food Calories = 1000 thermal calories

5 food Calories = 5000 calories = 5 kilocalories

Thus, when a piece of candy is totally burned, the thermal calorie presented in it will be released as heat.

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Eratosthenes determined the circumference of Earth by conducting an experiment. Put his steps in order as they correlate to the
galben [10]

Solution: The correct order is: C, A, B

The statement of the problem:

How can we prove Earth is round and calculate its circumference?

Hypotheis:

If the sun casts shadows at different angles at the same time of day in different places, we can determine how much Earth curves.

If the Earth was flat, the angle measured at different places at the same time of the day would be same.

Observation:

In Syene, the sun's rays are vertical at noon. At the same time in Alexandria, the rays are 7.2 degrees from the vertical.

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2 years ago
A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half. The density of the two smaller pieces is n
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A. Density only depends on the substance. It doesn't matter whether you have a little chip of it or a supertanker full of it ... the density doesn't change.
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2 years ago
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The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe
Novosadov [1.4K]

Answer:

a) a= 8.33 m/s²,    T = 12.495 N , b)    a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

            fr₁ + fr₂ = ma

The equation for the force of friction is

          fr = μ N

Y Axis

         N - W₁ –W₂ = 0

         N = W₁ + W₂

         N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

        N = 2 m g

We replace

           μ₁ 2mg + μ₂ mg = m a

          a = (2μ₁ + μ₂) g

          a = (2 0.30 + 0.25) 9.8

          a= 8.33 m/s²

Let's look for cable tension with beam 2

          T = m₂ a

          T = 1500 8.33

          T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

           fr = m₂ a₂

           N- w₂ = 0

          N = W₂ = mg

          μ₂ mg = m a₂

          a = μ₂ g

          a = 0.25 9.8

          a = 2.45 m / s²

4 0
2 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
2 years ago
Within an integrated circuit, each wafer is cut into sections, which
fiasKO [112]
The final answer is B hope its helps
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