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The cross product of two vectors is always perpendicular to both vectors. So, the magnetic force $\vec{F}=q\vec{v}\times\vec{B}$ will always be perpendicular to $\vec{v}$. This can be seen in your special case by realizing thatsince the motion is in the $(x,y)$ plane $v_z=0$, and the only component of $\vec{B}$ that is not null is $B_z=B$, we have ...

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Yes, you defined the zero of time as when the particle is at $(4,0,0)$. It passed through $(1,-1,5)$ one second before that.

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I set a coordinate system on the cone apex, with $+\hat{y}$ pointing up and $+\hat{x}$ pointing radially out at an azimouth (location) angle $\theta=0$. The position of the ball is defined by the distance from the apex up the slope to the ball $r$ and the longitude around the cone $\theta$ $$\vec{p} = {\rm Rot}(\hat{y}, \theta) \begin{pmatrix} r \sin \psi ... 0 There is a mathematical convention for direction angles \alpha in an x-y diagram: counter-clockwise from positive horizontal axis (your "east" direction). Since \tan \alpha = y/x is a periodic function with period \pi, \arctan (y/x) has two solutions in the range [0, 2\pi], while your calculator may give you just one solution, mostly in the ... 0 Typically in motion problems angles tend to be measured from the horizontal. For example, in projectile motion problems, the launch angle of the projectile is nearly always measured from the horizontal, and when doing force inclined-plane problems, the same is usually true. However, there are situations where that is not always true. Sometimes in statics ... 2 There is no hard and fast convention. What is important is that the direction you state is unambiguous. You might state a given angle as 20^\circ South of West, or 200^\circ counterclockwise from East. Neither is "more correct", but note that neither is ambiguous, either. 1 Consider for example, a plane vector and two orthogonal unit vectors \hat x and \hat y. Any vector in the plane can be expressed as$$\vec v = (\vec v \cdot \hat x) \;\hat x + (\vec v \cdot \hat y) \; \hat y = v_x\; \hat x + v_y\; \hat y$$So, you're correct, \vec b \cdot \hat a is the component of \vec b in the \hat a direction. And further, ... 1 The localized vector is a vector which we know its magnitude and direction and its point of application , which is the point where the force acts , if the force were free vector , how you will calculate for example the moment ? that won't make any sense ok the second principle i found it in Vector mechanics book called PRINCIPLE OF TRANSMISSIBILITY. check ... 0 For a force, you need the following bits of knowledge: The direction and magnitude of the force. The point it acts on. Some people call that pair of specifications a "localized vector", because it consists of both a location (#2) and a vector (#1). 0 Circular motion is a motion with acceleration, because the velocity vector is changed during the motion. Therefore the engine must be on all the time, and the resulting force is changing the rockets velocity vector's direction, but not size (under other circumstances than those give in the question, the rocket could be using this same propulsive force to ... 0 It is undergoing uniform circular motion, so its linear velocity is tangential to the circle, and its acceleration is towards the centre of the circle. You can easily prove that it is tangential using calculus on complex numbers; Let z=Re^{i \theta}=x+iy Differentiating both sides with respect to time, \frac{dz}{dt}=Ri e^{i \theta}*\omega=i \omega z, ... 0 As gravity force is their in Y direction so momentum is not conserved but in X direction as MC is valid so no change in the velocity. if take the components we can see the impulse is in positive Y direction only i.e. direction of force. so acceleration due to collision is in positive Y direction but gravity is their in negative Y direction. in X direction it ... 1 Normally, I think that the surface will react with a force in the OC→ direction Yes, if the ball's force on the surface is in direction A, then by Newton's third law, the surface's force on the wall is in direction C. This is what Newton's third law says. The third law applies to all forces in the same way. However, as pointed out by Jan Hudec in a ... 1 The force between two colliding bodies is not in the direction of motion. The normal and parallel components have to be treated separately. The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface ... 3 By the way, OA/OB/OC are not force vectors. The only force vector that is acting on the ball before (and after) it hits the ground is its weight, and it is acting vertically downwards. OA/OB/OC are velocity vectors. Try to split OA up into its vertical and horizontal components. When the ball hits the floor, reverse the vertical component due to impulse ... 2 Just use the definition http://en.wikipedia.org/wiki/Divergence \nabla\cdot F = r^{2n}+2nx^2r^{2(n-1)}+r^{2n}+2ny^2r^{2(n-1)}+r^{2n}+2nz^2r^{2(n-1)} = 3r^{2n}+2nr^{2n-1} 0 Assume the point mass to be at position \vec{r}. A spring at \vec{R_i} exerts a force along the line of connection.$$\vec{F_i} = -k_i (\vec{r}-\vec{R_i})$$Summing up and adding a term for gravity, it yields$$ \vec{F} = -\sum_i k_i (\vec{r}-\vec{R_i}) - m\vec{g}$$I can treat the N slinkys (I'm assuming they're parallel) as one large slinky. If ... 1 The x and y velocities should not add to V_0. To understand why, imagine something moving with V_x = 1 \frac{m}{s} and V_y = -1 \frac{m}{s}. This is something going horizontally and down; there's no reason why its velocity should be zero. The answer is that V_0 is the length of the velocity vector \vec{V}, and so it's calculated using ... 2 Momentum is a vector. For example, in 3D \mathbf{p}=(p_x,p_y,p_z). The magnitude of the momentum vector is a scalar: p=|\mathbf{p}|=\sqrt{p_x^2+p_y^2+p_z^2}. 0 To complete the given answers you have to keep in mind that the number of the cross product is the multiplication of the numbers of thus vectors and the sinus of the angle between them. If the angle is zero than there does not exist a direction of the cross product. 1 As with the other answers by Swami and Red ACt, the scalar product has no direction, the vector product is orthogonal to the plane defined by the two vectors (and nought if the two vectors are colinear). All this seems a bit arbitrary at first meeting: you're clearly a bit bewildered by this and this is quite normal. So, although the following is a bit ... 2 The result of a dot product is, as Red Act said, a scalar, it must not have a direction. The result of a cross product must be a vector. It must have a direction. Curl your right hand fingers - moving them from A towards B via the shorter angle, your thumb gives the direction of the result of the cross product. 3 The dot product of two vectors doesn't have a direction, because it's a scalar (single number), not a vector. The cross product A\times B is perpendicular to both A and B. If on your right hand, your thumb is sticking up and pointing in the direction of A, and your index finger is pointing in the direction of B, then your palm is facing in the ... 1 I'd be much obliged if someone could give me an example of such a field Consider a vector field \vec F with non-zero curl in the z direction only:$$\nabla \times \vec F = \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)\hat z $$An example of such a field is$$\vec F = -y \hat x + x \hat y$$which has curl ... 2 How picky are you about singularities? Consider the magnetic field of a steady current oriented in the +z direction. The field is$$\vec B \sim \frac{1}{r}\hat \phi, If you do a naive calculation of the curl, you'll find it's zero. And indeed, if you evaluate a closed line integral that does not bound $(x,y)=(0,0)$, you'll get zero. (Think of ...

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The answer is trivially positive. Take a conservative vector field $X$ in $\mathbb R^n$ and consider a closed smooth curve $\gamma$ in a open bounded region, say $\Omega\subset \mathbb R^n$. Obviously $\int_\gamma X \cdot d x =0$. Next consider another vector field $Y$ in $\mathbb R^n$ which is non-conservative in a similar region $\Omega'$ disconnected ...

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A vector doesn't have a head and a tail. A vector in this context is something with a magnitude, and except for the zero vector, a direction. Another way to look at it: The displacement vector from the origin to the point with coordinates (1,1,1) and the displacement vector from the point with coordinates (2,2,2) to the point with coordinates(3,3,3) are the ...

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You have to consider the head in order to know what direction the vector is in, right? After all, that's half the point of having a head - to let you know what direction the vector points. (But keep in mind that, except for position or displacement vectors, the position at which you draw the head doesn't mean anything. The vector itself is located at the ...

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