<h3>Hello there!</h3>
Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.
The equation commonly accepted to find the answer to questions like these is the specific heat formula.
The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.
The information given:
m = 187 grams
c = specific heat capacity of water, or in this case 1 calorie, or 4.184 joules (which is what we will be using)
ΔT = 80 degrees
Now just plug everything in to solve.
Q = 187 * 4.184 * 80
Q = 62592.64
So you have your answer: 62592.64 joules.
Hope this helped!
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
It doesn't matter. If the slides are truly frictionless, then
your kinetic energy at the bottom will be equal to the
potential energy you had at the top, no matter what kind
of route you took getting down.
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The only way I can think of that it would make a difference
would be if the shallow slide were REALLY REALLY long,
and you didn't have anything to eat all the way down.
Then you might lose some weight while you're on the slide,
and your mass might be less at the bottom than it was at the
top. Then, in order to have the same kinetic energy at the
bottom, you'd need to be going a little bit faster.
But if it takes less than, say, two or three days, to go down the
long, shallow slide, then this effect would probably be too small
to make any difference.
The answer is true because A current carrying wire is surrounded by magnetic field
To demonstrate the number of protons neutrons and electrons.
