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:
Explanation:
Given equation is ,
x =t + 2 t³ ,
dx/dt = velocity ( v ) = 1 + 6 t²
a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²
b ) Acceleration = dv /dt = 12 t .
force( F ) = mass x acceleration = 4 x 12 t = 48 t
Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )
work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; = [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² = 864 J
Answer:
The average force on ball by the golf club is 340 N.
Explanation:
Given that,
Mass of the golf ball, m = 0.03 kg
Initial speed of the ball, u = 0
Final speed of the ball, v = 34 m/s
Time of contact, 
We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

So, the average force on ball by the golf club is 340 N.
Answer:
1.41 m/s^2
Explanation:
First of all, let's convert the two speeds from km/h to m/s:


Now we find the centripetal acceleration which is given by

where
v = 12.8 m/s is the speed
r = 140 m is the radius of the curve
Substituting values, we find

we also have a tangential acceleration, which is given by

where
t = 17.0 s
Substituting values,

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:
