At the player's maximum height, their velocity is 0. Recall that
which tells us the player's initial velocity 
is
The player's height at time 
is given by
so we find their airtime to be
The radiation is ultra voilet or Gamma radiation , because their wave length is very short i e 1..0 to 2.5 (angstrom)Ao.
True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
Answer:
Since the maximum thermal efficiency is higher than 55 percent, there can be a power cycle with these reservoir temperature with an efficiency higher than 55 percent.
Explanation:
The maximum thermal efficiency is determined from the given temperature
nth Carnot = 1- TL/TH
Where TL= 17+273= 290k
TH= 627*273= 900K.
nth Carnot = 1- 290/900 = 0.68
0.68*100 = 68 percent
