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11Alexandr11 [23.1K]
2 years ago
8

the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc

h energy did the stove supply to melt the cookie sheet?
Physics
1 answer:
sergejj [24]2 years ago
3 0

Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

  • Heat required = mass * heat capacity * temperature change

Assuming the aluminum sheet was at room temperature initially.;

Room temperature = 25 °C

Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

Learn more about heat capacity at: brainly.com/question/21406849

#SPJ1

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The total electric field consists of the vector sum of two parts. One part has a magnitude of E1 = 1300 N/C and points at an ang
V125BC [204]

Answer:

2954.6 N/C, 46.36 degree from positive  axis

Explanation:

E1 = 1300 N/C, θ1 = 35 degree

E2 = 1700 N/C, θ2 = 55 degree

Now write the electric fields in vector form

E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j

E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j

Resultant electric field

E = E1 + E2

E =  1064.9 i + 745.6 j + 975.08 i + 1392.6 j

E = 2039.08 i + 2138.2 j

Magnitude of E

E = sqrt (2039.08^2 + 2138.2^2)

E = 2954.6 N/C

Let it makes an angle Φ from X axis

tan Φ = 2138.2 / 2039.08 = 1.049

Φ = 46.36 degree from positive X axis.

5 0
3 years ago
What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total m
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Answer:   
 M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b] 
 Î”R is the difference in density between the gas inside and surrounding the balloon. 
 R[b] is the density of gas inside the baloon.   
 ==================================== 
 Let V be the volume of helium required. 
 Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V   
 U = 1.225gV newtons 
 ---- 
 Weight of Helium = Volume of Helium * Density of Helium * g 
 W[h] = 0.18gV N   
 Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N  -----

 
 Weight of 260kg = 2549.7 N 
 Then to lift the whole thing, F > 2549.7 
 So minimal F would be 2549.7 
 ---- 
 1.045gV = 2549.7 
 V = 248.8 m^3   
 Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)   
 =====   
 Let the density of the surroundings be R 
 Then U-W = (1-0.9)RgV = 0.1RgV   
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3 years ago
Do you think the plan for the skyscraper in California is appropriate based on its location? Why or why not? Your answer should
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Answer:

No

Explanation:

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What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?
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Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

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