Answer:
The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radius 3 centimeters at the bottom of a cylindrical tank of radius 10 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the acceleration of gravity, which you should approximate as 1000 cm/s2.
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Answer:
The magnitude of the lift force L = 92.12 kN
The required angle is ≅ 16.35°
Explanation:
From the given information:
mass of the airplane = 9010 kg
radius of the airplane R = 9.77 mi
period T = 0.129 hours = (0.129 × 3600) secs
= 464.4 secs
The angular speed can be determined by using the expression:
ω = 2π / T
ω = 2 π/ 464.4
ω = 0.01353 rad/sec
The direction 

θ = 16.35°
The magnitude of the lift force L = mg ÷ Cos(θ)
L = (9010 × 9.81) ÷ Cos(16.35)
L = 88388.1 ÷ 0.9596
L = 92109.32 N
L = 92.12 kN
Calcium chloride contains ionic bonds.
Pennies contain metallic bonds.
Hydrochloric acid contains covalent bonds.
You're welcome.
Vt = Vboat - Vriver
Vt = 18 - 2.5 = 15.5 m/s
If the boat's direction is the same as the water, you sum the velocities of the river and the boat .
Answer:
24m/s²
Explanation:
Given
Distance S = 3m
Time of fall = 0.5sec
Required
Acceleration due to gravity
Using the equation of motion
S = ut+1/2gt²
Substitute the given values
3 = 0+1/2g(0.5)²
3 = 1/2(0.25)g
3 = 0.125g
g = 3/0.125
g = 24
Hence the value for the acceleration of gravity on this new planet is 24m/s²