Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity 
= 60 mph = 26.8224 m/s
Final velocity 
= 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = 
m( ² - ² )
we substitute
Δk = ×1814.37( (26.8224)² - (13.4112)² )
Δk = × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Answer:
pressure, stress pascal N/m2
energy, work, quantity of heat joule N·m
power, radiant flux watt J/s
electric charge, quantity of electricity coulomb -
A) It will be 2 covalent bonds
B) covalent bonds occur when there’s 2 atoms that share electrons. In this case by sharing the 2 pairs of valence electrons each atom has a total of 8 valence electrons
Use pythagorean's theorem for this, with 7 as a and 5 as b. pythagorean's theorem says that a^2 + b^2 = c^2, so 7^2 * 5^2 = c^2. this gives you 49 + 25 = c^2, so 74 = c^2. c = sqrt 74, which is approximately 8.60 km
