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Step Formula / Calculation Reasoning
1 \text{Total initial speed} = \sqrt{v_{\text{initial}}^2 + v_{\text{vertical initial}}^2} Calculating the total initial speed of the fish combining its horizontal and vertical velocities.
2 \text{Total initial speed} = \sqrt{6.0^2 + 0^2} = 6.0 , \text{m/s} Initial vertical velocity is zero as the fish is dropped, not thrown downwards.
3 v_{\text{vertical double}} = \sqrt{\text{double speed}^2 – v_{\text{initial}}^2} Calculating the vertical component of velocity when the total speed doubles.
4 \text{Double speed} = 2 \times \text{total initial speed} = 12.0 , \text{m/s} Determining the speed value that is double the initial speed.
5 v_{\text{vertical double}} = \sqrt{12.0^2 – 6.0^2} Finding the vertical component when total speed is 12.0 m/s.
6 t_{\text{double}} = \frac{v_{\text{vertical double}} – v_{\text{vertical initial}}}{g} Time calculation for the speed to double using the kinematic equation.
7 t_{\text{double}} \approx 1.06 s Evaluating the time for the fish’s speed to double.
8 v_{\text{vertical quadruple}} = \sqrt{\text{quadruple speed}^2 – v_{\text{initial}}^2} Calculating the vertical component of velocity when the total speed quadruples.
9 \text{Quadruple speed} = 2 \times \text{double speed} = 24.0 , \text{m/s} Determining the speed value that is double the double speed.
10 v_{\text{vertical quadruple}} = \sqrt{24.0^2 – 6.0^2} Finding the vertical component when total speed is 24.0 m/s.
12 t_{\text{quadruple}} \approx 2.37 s Evaluating the time for the fish’s speed to quadruple.
13 \text{Additional time} = t_{\text{quadruple}} – t_{\text{double}} Calculating the additional time required for the fish’s speed to double again.
14 \text{Additional time} \approx 1.31 s Evaluating the additional time needed for the fish’s speed to double again from the double speed.

The time for the fish’s speed to double is approximately 1.06 seconds. The additional time required for the fish’s speed to double again (from double to quadruple the initial speed) is approximately 1.31 seconds.

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1. 1.06 s
2. 1.31 s

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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