Kinetic energy = (1/2) (mass) x (speed)²
At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules
At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules
The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.
That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
Answer
Together with glycolysis, The Krebs cycle, and the electron transport chain release about 36 molecules of ATP per molecule of glucose.The Krebs cycle uses the two molecules of pyruvic acid formed in glycolysis and yields high-energy molecules of NADH and flavin adenine dinucleotide (FADH2), as well as some ATP. The electron transport chain forms a proton gradient across the inner mitochondrial membrane, which drives the synthesis of ATP
Given:
Time: 3.5 hrs
Velocity: 120 miles/hr
Now Distance= Speed × Time
Now Velocity and speed have the same magnitude. Velocity being a vector quantity has a definite direction. Whereas speed is a scalar quantity,it indicates only the magnitude an doesn't define any direction.
Hence Distance = Velocity x time
Distance = 3.5 × 120 = 420 miles
Answer: 1.176×10^-3 s
Explanation: The time constant formulae for an RC circuit is given below as
t =RC
Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F
t = 56×10^-6 × 21
t = 1176×10^-6
t = 1.176×10^-3 s
Answer:
B) 18,000 feet MSL
Explanation:
There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.
