Answer:
ΔK = -6 10⁴ J
Explanation:
This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved
initial instant. Before the crash
p₀ = m v₁ + M 0
final instant. Right after the crash
p_f = (m + M) v
p₀ = p_f
mv₁ = (m + M) v
v =
we substitute
v =
3
v = 1.0 m / s
having the initial and final velocities, let's find the kinetic energy
K₀ = ½ m v₁² + 0
K₀ = ½ 20 10³ 3²
K₀ = 9 10⁴ J
K_f = ½ (m + M) v²
K_f = ½ (20 +40) 10³ 1²
K_f = 3 10⁴ J
the change in energy is
ΔK = K_f - K₀
ΔK = (3 - 9) 10⁴
ΔK = -6 10⁴ J
The negative sign indicates that the energy is ranked in another type of energy
A) The power delivered to the lines is

And the voltage at which the lines work is

Since the power delivered is the product between the voltage and the current:

We can find the current flowing in the lines:

b) The voltage change along each line can be found by using Ohm's law:

c) The power wasted as heat along each line is given by:

And since we have 2 lines, the total power wasted as heat in both lines is
Not sure if this it completely right but here’s what I got: 77.16050617284 J
A star’s death also depends on its mass. The most massive stars
quickly exhaust their fuel supply and explode in core-collapse
supernovae, some of the most energetic explosions in the universe. A
supernova’s radiation can easily (if only briefly) outshine the rest of
its host galaxy. The remnant stellar core will form a neutron star
or a black hole, depending on how much mass remains. If the core
contains between 1.44 and 3 solar masses, that mass will crush into a
volume just 10 to 15 miles wide before a quantum mechanical effect known
as neutron degeneracy pressure prevents total collapse. The
exact upper limit on a neutron star mass isn’t known, but around 3 solar
masses, not even neutron degeneracy pressure can combat gravity’s
inward crush, and the core collapses to form a black hole.
Answer:
CaCl2 is a reactant
Explanation:
Calcium carbonate (CaCO3) is a reagent, it is found on the left side, what is found on the right side are the products.