- #1
yashboi123
- 17
- 0
- Homework Statement
- What is the cross product of A X C?
- Relevant Equations
- A X B = ABSin(x)
Why are they using Cosine instead of Sine for Cross Product?
- Thread starter yashboi123
- Start date
AI Thread Summary
The discussion centers on the use of cosine instead of sine for calculating the cross product, specifically addressing the negative sign in the answer and the use of a 25-degree angle. The negative sign is explained through the right-hand rule, indicating that the direction of the vectors leads to a negative result when following the anti-clockwise rotation. The angle AOC is derived as 90 degrees minus the marked 25-degree angle, which leads to the use of cosine in the calculations. Additionally, the relationship between sine and cosine is clarified, showing that sin(295) equals -cos(25), reinforcing the choice of cosine in the context. Understanding these trigonometric relationships is crucial for accurate vector calculations in physics.
- #2
- 4,140
- 1,741
The answer is -96 times sine of angle AOC.
The minus sign arises from the right-hand rule. Take your right hand, point the thumb in the direction of the positive z axis, and curl your fingers. The direction your fingers curl, which is anti-clockwise as we look at the diagram, must be the direction from the vector that's the first argument to the cross product to the vector that's the second argument, in order to get a positive sign. Since in this diagram ##\vec A## is clockwise from ##\vec C## we get a negative sign.
Now what about that cos? The diagram does not mark angle AOC, but we know it it is 90 degrees minus the marked 25 degree angle.
So they just use the formula for sine of the sum of two angles, as follows:
\begin{align*}
\sin\ AOC &= \sin(90 - 25) \\&= \sin(90 + (-25))\\& = \sin 90\ \cos (-25) + \cos 90\ \sin(-25)
\\&= \cos(-25)\times 1 + 0\times \sin(-25)
\\&= \cos\ 25+0
\\&= \cos\ 25\end{align*}
The minus sign arises from the right-hand rule. Take your right hand, point the thumb in the direction of the positive z axis, and curl your fingers. The direction your fingers curl, which is anti-clockwise as we look at the diagram, must be the direction from the vector that's the first argument to the cross product to the vector that's the second argument, in order to get a positive sign. Since in this diagram ##\vec A## is clockwise from ##\vec C## we get a negative sign.
Now what about that cos? The diagram does not mark angle AOC, but we know it it is 90 degrees minus the marked 25 degree angle.
So they just use the formula for sine of the sum of two angles, as follows:
\begin{align*}
\sin\ AOC &= \sin(90 - 25) \\&= \sin(90 + (-25))\\& = \sin 90\ \cos (-25) + \cos 90\ \sin(-25)
\\&= \cos(-25)\times 1 + 0\times \sin(-25)
\\&= \cos\ 25+0
\\&= \cos\ 25\end{align*}
- #3
- 42,656
- 10,435
@andrewkirk , I would think @yashboi123 is looking for what is wrong with sin(295), following a standard method, rather than for an alternate method. Which method the book used is unknown.
There are some useful formulas:
cos(90-x)=sin(x)=sin(180-x)=-sin(-x)=sin(360+x)
So sin(295)=sin(360-65)=sin(-65)=-sin(65)=-cos(25).
As your calculator will tell you, sin(295)=-cos(25).yashboi123 said:
There are some useful formulas:
cos(90-x)=sin(x)=sin(180-x)=-sin(-x)=sin(360+x)
So sin(295)=sin(360-65)=sin(-65)=-sin(65)=-cos(25).
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are;
Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction.
Must be LESS than 1 kg.
NO parachutes.
NO glue or Tape can touch the egg.
MUST be able to take egg out in less than 1 minute.
Grade A large eggs will be used.
Similar threads
Hot Threads
-
Torque biomechanics
- Started by Naucus
- Replies: 6
- Introductory Physics Homework Help
-
Work Done in slowly pulling a thread
- Started by Bling Fizikst
- Replies: 28
- Introductory Physics Homework Help
-
How do you calculate the pull of this piston?
- Started by vdance
- Replies: 27
- Introductory Physics Homework Help
Recent Insights
-
Insights Why Entangled Photon-Polarization Qubits Violate Bell’s Inequality
- Started by Greg Bernhardt
- Replies: 7
- Quantum Physics
-
Insights Quantum Entanglement is a Kinematic Fact, not a Dynamical Effect
- Started by Greg Bernhardt
- Replies: 11
- Quantum Physics
-
Insights What Exactly is Dirac’s Delta Function? - Insight
- Started by Greg Bernhardt
- Replies: 5
- General Math
-
Insights Relativator (Circular Slide-Rule): Simulated with Desmos - Insight
- Started by Greg Bernhardt
- Replies: 1
- Special and General Relativity
-
Insights Fixing Things Which Can Go Wrong With Complex Numbers
- Started by PAllen
- Replies: 11
- General Math
-
Insights Fermat's Last Theorem
- Started by fresh_42
- Replies: 105
- General Math