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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 ...


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

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.


2

Just use the definition http://en.wikipedia.org/wiki/Divergence $\nabla\cdot F = r^{2n}+2nx^2r^{2(n-1)}+r^{2n}+2ny^2r^{2(n-1)}+r^{2n}+2nz^2r^{2(n-1)} = 3r^{2n}+2nr^{2n-1}$


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.


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: $p=|\mathbf{p}|=\sqrt{p_x^2+p_y^2+p_z^2}$.


2

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 ...


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 ...


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 ...


1

The formula you have written is correct; but they are functions of time. Hence, by inserting the particular instant , say $t$ on the function ,you get the instantaneous components of velocity. Then using phythagoras theorem you will get the total instantaneous velocity. Taking your example, at time $T$ s , the X-comp. is $30$ unit and Y-comp. is $(20 - ...


1

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 ...


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 ...


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 ...


1

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 ...


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 ...


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 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 ...


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, ...



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