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Sophie [7]
3 years ago
6

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

Physics
1 answer:
Andreyy893 years ago
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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HELP PLEASE...
skad [1K]
Ya because they’re are both 50 liters
7 0
3 years ago
Accomplished silver workers in india can pound silver into incredibly thin sheets, as thin as 3.00 10-7 m (about one-hundredth o
postnew [5]
The density of silver is ρ = 10500 kg/m³ approximately.

Given:
m = 1.70 kg, the mass of silver
t = 3.0 x 10⁻⁷ m, the thickness of the sheet

Let A be the area.
Then, by definition,
m = (t*A)*ρ

Therefore
A = m/(t*ρ)
    = (1.7 kg)/ [(3.0 x 10⁻⁷ m)*(10500 kg/m³)]
    = 539.7 m²

Answer: 539.7 m²

8 0
3 years ago
The chemical equation provided shows iron rusting to form iron oxide. Use the drop-down menu to choose the coefficients that wil
Brrunno [24]

Answer:

4 Fe + 3 O_2 \rightarrow 2 Fe_2 O_3

Explanation:

The original equation is:

Fe + O_2 \rightarrow Fe_2 O_3

We notice that:

- we have 1 atom of Fe on the left, and 2 atoms of Fe on the right

- we have 2 atoms of O on the left, and 3 atoms of O on the right

Therefore, the equation is not balanced.

In order to balance it, we can add:

- a coefficient 3 in front of O_2

- a coefficient 2 in front of Fe_2 O_3

So we have:

Fe + 3 O_2 \rightarrow 2Fe_2O_3

Now the oxygen is balanced, but the iron it not balanced yet, since we have 1 Fe on the left and 4 on the right. Therefore, we should add a coefficient 4 on the Fe on the left:

4 Fe + 3 O_2 \rightarrow 2 Fe_2 O_3

3 0
2 years ago
Read 2 more answers
A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the b
jonny [76]

Answer

given,

vertical speed of stone,v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0  

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

   v = u + a t

   v = 12 - 9.8 x 5.2

  v = -38.96 m/s

ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

c) total distance travel by the stone

  vertical distance covered by the stone

 v² = u² + 2 g h

 0 = 12² - 2 x 9.8 x h

 h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.

8 0
3 years ago
An electron is projected with an initial speed Vo = 5.35x10^6 m/s into the uniform field between the parallel plates. The direct
mart [117]

Answer:

 E = 3.04 10⁻⁵ N / C

Explanation:

In this problem we can use the kinematics to find how long it takes the electron to travel the plates

Let's start by reducing the magnitudes to the SI system

              vₓ = 5.35 10⁶ m / s

              x = 2 cm = 2 10⁻² m

              y = 1 cm = 1 10⁻² m

             x = vₓ t

             t = x / vₓ

             t = 2 10⁻² / 5.35 10⁶

             t = 3,738 10⁻⁹ s

This time is also the time it takes for vertical movement to go from the center to the plate, let's look for acceleration with Newton's second law

              F = m a

              a = F / m = e E / m

              y = v_{oy} + ½ a t²

              v_{oy} = 0

We replace

              y = ½ e / m E t²

              E = 2 y m / e t²

Let's calculate

            E = 2  1 10⁻²  9.1 10⁻³¹ / (1.6 10⁻¹⁹  3,738 10⁻⁹)

            E = 18.2 10⁻³³ / 5.98 10⁻²⁸

            E = 3.04 10⁻⁵ N / C

5 0
3 years ago
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