made up of more than one cell.
The bicyclist accelerates with magnitude <em>a</em> such that
25.0 m = 1/2 <em>a</em> (4.90 s)²
Solve for <em>a</em> :
<em>a</em> = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²
Then her final speed is <em>v</em> such that
<em>v</em> ² - 0² = 2<em>a</em> (25.0 m)
Solve for <em>v</em> :
<em>v</em> = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s
Convert to mph. If you know that 1 m ≈ 3.28 ft, then
(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h
Answer:
The radius of the gold nucleus is 7.1x10⁻¹⁵m
Explanation:
The nearest distance is:
(eq. 1)
Where
z = atomic number of gold = 79
e = electron charge = 1.6x10⁻¹⁹C
k = electrostatic constant = 9x10⁹Nm²C²
energy of the particle = 32 MeV = 5.12x10⁻¹²J
At the potential energy is zero, all the energy will be kinetic energy:

Where
m = 4 mp = mass of proton

Replacing in equation 1

Kinetic energy of golf club = 65J,
kinetic energy supplied to golf ball = 20% of 65 = 0.2 * 65 = 13J,
kinetic energy of ball = [mass * Velocity²]/2,
mass = 46gm = 0.046Kg,
[0.046 * V²]/2 = 13, or 0.046 *V² = 26,
V² = 26/0.046 = 565.22,
V = 23.77 m/sec = initial velocity of golf ball after hitting.
Answer:
A) The speed of the water must be 8.30 m/s.
B) Total kinetic energy created by this maneuver is 70.12 Joules.
Explanation:
A) Mass of squid with water = 6.50 kg
Mass of water in squid cavuty = 1.55 kg
Mass of squid = 
Velocity achieved by squid = 
Momentum gained by squid = 
Mass of water = 
Velocity by which water was released by squid = 
Momentum gained by water but in opposite direction = 
P = P'


B) Kinetic energy does the squid create by this maneuver:
Kinetic energy of squid = K.E =
Kinetic energy of water = K.E' = 
Total kinetic energy created by this maneuver:

