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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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What is the momentum of a 1550kg car that is traveling 38.0 m/s?
konstantin123 [22]

Answer:

p = 58,900 kg m/s

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p = m × v

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A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?
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U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
                   =(30-10)/6
                   =(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
                              =60+ 0.5*(10/3)*36
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And you can solve it in another way:

v^2=u^2+2as
or, s=(v^2-u^2)/2a
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A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
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I attached the missing picture.
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There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
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The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\&#10;mg=m\frac{v^2}{r}\\&#10;gr=v^2\\&#10;v=\sqrt{gr}=8.29\frac{m}{s}

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