To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,
Here,
= Mass of Aluminum
= Specific Heat of Aluminum
= Specific Heat of Water
Mass of water
Latent of Vaporization
Replacing,
Converting,
Therefore is required 440.409kCal
Answer:
60m/s
Explanation:
initial energy = final energy
g.p.e = k.e
k.e = 0.5 × mass × velocity²
g.p.e = 990000J as per Question
990000Nm = 0.5 × 550 × V²
V² = 3600
V = 60m/s
In general, all views of the cosmic microwave background are identical. Isotropy is demonstrated by this.
<h3>What exactly is isotropy, for instance?</h3>
The Greek words isos (equal) and tropos, from which the term "isotropy" is derived, mean "uniform in all directions" (way). The material properties of anisotropic materials, such as graphite, differ depending on the direction, in contrast to isotropic materials like glass, which show the same properties in all directions.
There is no "centre" to an isotropic universe, which is another characteristic. The North and South Poles are produced by the rotation of the Earth, giving them a distinctive orientation, but the Universe is visible from every angle. When we think about the Big Bang, which is the origin of the Universe, this is a crucial point.
To know more about cosmic microwave background visit:
brainly.com/question/28197648
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Answer:
C) No work is required to move the negative charge from point A to point B.
Explanation:
An equipotential surface is defined as a surface connecting all the points at the same potential.
Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.
The work done when moving a charge is given by
where
q is the charge
is the potential difference between the initial and final point of motion of the charge
However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so
And so, the work done is also zero.
I believe that B is the answer.
