Answer:
Magnets exert forces and torques on each other due to the rules of electromagnetism. The forces of attraction field of magnets are due to microscopic currents of electrically charged electrons orbiting nuclei and the intrinsic magnetism of fundamental particles (such as electrons) that make up the material. Hope this helps you! :)
Answer:
Option 3: -48 cm
Explanation:
We are given:
refractive index; n = 1.5
radius of curvature; r2 = 24 cm
Formula for the focal length is given as;
1/f = (n - 1) × [(1/r1) - (1/r2)]
As r1 tends to infinity, 1/r1 = 0
Thus,we now have;
1/f = (n - 1) × (-1/r2)
Plugging in the relevant values;
1/f = (1.5 - 1) × (-1/24)
1/f = -0.02083333333
f = -1/0.02083333333
f = -48 cm
Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:

where

giving

The rotational part requires the moment of inertia of a solid cylinder

Then the rotational kinetic energy is

Adding the two types of energy and factoring out common terms gives

Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.
The answer is D.Puck.
♡♡Hope I helped!!! :)♡♡
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
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