Answer:
A) attached file
B) attached file
C) attached file
D) Kirchhoff’s junction rule states that at any junction, the sum of the altimeter attained moving into and out of that junction are equal.
While
Kirchhoff’s loop rule states that the algebraic sum of the number of lifts used in any closed loop is equal to zero
Explanation:
Given that the lifts are analogous to batteries, and the runs are analogous to resistors.
So from all the figures. The resistors represent the runs while the lift represents the battery.
Kirchhoff’s junction rule states that at any junction, the sum of the altimeter attained moving into and out of that junction are equal.
While
Kirchhoff’s loop rule states that the algebraic sum of the number of lifts used in any closed loop is equal to zero
Please find the attached file for the sketch
Answer:
2.74 × 10^33 J
Explanation:
the formula to calculate kinetic energy is:
1/2mv²
m= mass (kg)
v= velocity (m/s)
given that,
m = 5.97 × 10^24
v = 30.29 km s-1
= 30290 m s-1
1/2× 5.97 × 10^24 × 30290²
=2.74 × 10^33 J
The number of protons in an atom is the same as its atomic number, which is found on its corresponding atomic symbol in the periodic table of elements. And as long as this atom has a neutral charge, than the number of electrons will be the same. If the atom does have a charge, like say, 2-, then subtract 2 from the number of protons to get the number of electrons.
For the number of neutrons, you have to memorize the isotopes available sometimes. But usually something like carbon 14 will have a total of (in this case) 14 total protons/neutrons. So subtract this number by the number of protons(6), and you have 8 neutrons.
Hope this helps!
Answer:
1.15×10⁶ J
Explanation:
Heat needed to bring the ice to 0°C:
q = mCΔT
q = (3 kg) (2100 J/kg/K) (0°C − (-2°C))
q = 12,600 J
Heat needed to melt the ice:
q = mL
q = (3 kg) (3.36×10⁵ J/kg)
q = 1,008,000 J
Heat needed to bring the water to 10°C:
q = mCΔT
q = (3 kg) (4200 J/kg/K) (10°C − 0°C)
q = 126,000 J
The total heat is:
q = 12,600 J + 1,008,000 J + 126,000 J
q = 1,146,600 J
q = 1.15×10⁶ J
Full Question:
Consider an automobile with a mass of 5510 lbm braking to a stop from a speed of 60.0 mph.
How much energy (Btu) is dissipated as heat by the friction of the braking process?
______Entry field with incorrect answer Btu
Suppose that throughout the United States, 350.0 x 10^6 such braking processes occur in the course of a given day.
Calculate the average rate (megawatts) at which energy is being dissipated by the resulting friction.
______Entry field with incorrect answer MW
Answer:
Energy dissipated as heat = 852.10 Btu
Average rate of energy dissipation = 3.64 MW
Explanation:
1 lb = 0.453592 Kg
Mass = 5510 lb = 0.453592 * 5510 Kg = 2499.29 Kg
1 mph = 0.44704 m/s
Velocity = 60 mph = 60 * 0.44704 m/s = 26.82 m/s
Let E be equal to energy acquired
Energy dissipated as heat:
E = 852.10 Btu
Energy dissipated throughout the United States
Average power rate in MW, P
