Unusual precipitation patterns
Answer:
Energy expenditure in K cals/min = 10 K cals /min (approximately)
Explanation:
As we know
Energy expenditure in Kcal/min= METs x 3.5 x Body weight (kg) / 200
Given is METs=7.6
Weight of Jazz= 172lb=78.02kg
putting the values in formula,
Energy expenditure in K cals/min= 7.6 x 3.5 x 78.02 / 200
=10.38 K cals /min
=10 K cals /min (approximately)
Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min
Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
We are given with
distance traveled through vacuum = 1.0 m
refractive index of water = 1.33
refractive index of glass = 1.50
refractive index of diamond = 2.42
distance traveled through water is = 1.0/1.33 = 0.75 m
distance traveled through water is = 1.0/1.50 = 0.67 m
distance traveled through water is = 1.0/2.42 = 0.41 m
Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°
