# Tag Info

## New answers tagged vectors

3

You can actually infer the difference of the approaches just by looking at their names. A vector has direction and an energy (a scalar quantity) does not. Therefore, when you are trying to figure out scalar quantities such as distance and speed, you may find energy method more advantageous; when you look for velocity, acceleration, you have to use vector ...

0

The ball is rolling along positive x direction with velocity v. The points of the ball directly below the center are rotating around the center of the ball in the -x direction with instantaneous velocity u equal to the distance from the center of the ball times the angular velocity of the ball. Where v = u you have the condition you are looking for. Truly ...

2

Torque $\vec{\tau}(\vec{r})$ is an vector multiplication of radius-vector $\vec{r}$ on applied force vector $\vec{F}$, i.e. $\vec{\tau}=[\vec{r}\times\vec{F}]$ Here the radius-vector (or position vector) $\vec{r}$ is the vector from the point where the torque is defined to the point where the force is applied (see image). On the picture you shown ...

0

The length or distance is actually the relative position vector taken from some fixed reference point. That reference point in your example is most conveniently taken to be the center of the bolt. Specifying distance is not enough. You need also the angle between the applied force and the position vector. Distance alone does not give you what you need to ...

1

I think this can be best understood by an example. The distance between two cities is a scalar quantity, say 500 km. If you want to know how much gas you are going to burn when you drive from one city to other, the distance is the thing you want to know. However position of a city with respect to another one (or with respect to any point in space) is a ...

0

First of all, let us build a coordinate system where +x is east, -x is west, +y is north, -y is south. So your initial velocity $v_i = (-22, 0)$ Your final speed is 12 m/s, but since you are traveling in the South West Direction, you need to use Pythagorean Theorem to find the vector components in the x-y coordinate system. 12 is the hypotenuse, so the legs ...

2

Physicists tend to be a bit casual about sign conventions when it seems to be obvious. So let's attempt to be completely rigourous. The key step is getting the flight time $t$ since the range is just $v\cos\theta\, t$. We do this using the SUVAT equation: $$v = u + at$$ We'll use the usual conventions that up and right are positive, so $v_y$ and $v_x$ ...

3

Well, a good example is thinking in term of components. In several areas of physics, the math gets more intuitive when you think in terms of components of the vectors. So, instead of writing the vector $\mathbf r$ for the position of a particle, you write $x^i$ as the $i$-th component of a vector. The $i$ in the top is to indicate a contravariant vector, ...

-2

***the 2N force was cancelled because of the force acting horizontally which i call frictional Force............but when the same situation is carried in a interstellar space free from gravity and other forces the mass will experience a total of 7N force because there is no friction to cancel the resultant forces.

1

Your proof is entirely correct. Relativity doesn't have to be difficult :) To be clear, the steps in your proof are simply the definition of $U'$, the definition of $X'$, the product rule, the constancy of $\Lambda$, and the definition of $U$, respectively. There is nothing wrong with any of these steps. As for why $\tau$ and not $t$: $\tau$ is defined ...

2

I've tried to do it the following way, but I don't know if I can use the product rule as such when matrices and vectors are involved. \begin{equation*} U'=\frac{dX'}{d\tau}=\frac{d}{d\tau}\Lambda{X}=\frac{d\Lambda}{d\tau}X+\Lambda\frac{dX}{d\tau}=\Lambda\frac{dX}{d\tau}=\Lambda{U} \end{equation*} since the Lorentz transformation matrix ...

1

There are two different questions that are unrelated to each other. The first one is how the 4-velocity is defined. By definition, velocities are tangent flows on a differential manifold, therefore derivatives must be taken with respect to the parameter you are using to describe the flow with. In the context of special relativity such parameter is the ...

1

The modern notions that separate "scalars" and "vectors" goes as follows: Scalars are elements of fields. Examples of fields include the rational numbers, the real numbers, and the complex numbers. Scalars can be added and multiplied and divided. Vectors are spaces over fields. That is, lists of elements of fields. Velocity vectors, for example, are ...

1

Think of a vector as having direction in space (north, south, east, west). Scalars may or may not be capable of having negative values. It just depends on the nature of the quantity. The statement that negative values for scalars are just convention is rather misleading. Some "conventions" just naturally make a whole lot of sense, and changing them would ...

1

Start by rewriting the scalar product as a covariant-contravariant contraction, like so: $${\bf u}\cdot{\bf v} = g_{ij}u^iv^j = (g_{ij}u^i)v^j = u_jv^j$$ Now transform the components with your $S$ and $T$ matrices, $$u_jv^j = \left( S_j^a {\bar u}_a \right) \left( T^j_b {\bar v}^b \right) = (S_j^a T^j_b) {\bar u}_a {\bar v}^b = \delta^a_b {\bar u}_a {\bar ... 0 It has to do with what point you define as a reference, what do you call (define) as zero? When talking about temperature it depends which unit you measure in. In example, when using Celsius, zero is defined as the freezing / melting point of water (under normal pressure etc). Do we know anything colder than that? Yes we do. The only way to expres these ... 2 The sign of a scalar depends upon the scale with which it is measured. In the case of temperature, a Fahrenheit scale arbitrarily says 32 degrees is the freezing point of water, and zero degrees is a mark on a scale of numbers. Any temperature less than zero has a negative sign. The Celsius scale says zero is the freezing point of water, so temperatures ... -2 I think it should have dimensions. Suppose the question were "what is the unit vector of 10 Newtons force pointing due north?". Then the answer is "1 Newton due north". 1 \frac{1}{2}\partial_iu_ju_j is more clearly written \frac{1}{2}\partial_i(u_ju_j) which can be evaluated by the chain rule or the product rule. Using the product rule you get \frac{1}{2}(u_j\partial_iu_j+u_j\partial_iu_j), and using the chain rule you get \frac{1}{2}(2u_j\partial_iu_j). And technically you need linearity since you want to do this ... 4 Review of vectors A vector is a quantity with a magnitude and a direction, which makes it somewhat different from how normal quantities, which are just magnitudes. One huge difference is that vectors "transform" a certain way when we rotate our coordinates, for example. They "transform" just the same way that an arrow in space transforms, if we do not care ... 3 I find this type of question is always easier if you draw a diagram. The drag force F acts in the opposite direction to the velocity so it looks like: and the components of the drag force are:$$\begin{align} F_x &= F \cos\theta \\ F_y &= F \sin\theta \end{align}$$\cos\theta is v_x/v and v = \sqrt{v_x^2 + v_y^2} so we get:$$ F_x = F ...

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