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a) The reason its $F=mg\sin\theta$ is because that is how much of the weight is pointing down the slope. This makes since since when $\theta=0$ we would have to apply $0N$ so that it doesn't slide down (or remain at constant speed), and at $\theta=90$ we would have to apply $mg$ to keep it from sliding down (essentially hold the entire mass). b) Same idea ...

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The canoe has a relative velocity to the river which we simply call $\vec{v}_{c/r} =( \dot{x}, \dot{y} )$. The motion of the river relative to the earth is $\vec{v}_{r/e} = ( 0.54, 0 )$. The canoe relative to the earth is $\vec{v}_{c/e} = ( 0.55\cos(-45^\circ), 0.55\sin(-45^\circ) )$. All together you have $$\vec{v}_{c/e} = \vec{v}_{c/r} + \vec{v}_{r/e} ... 0 You've done the "most difficult" which is to get as many independant equations as needed to solve for the unknowns. The first one flows from the fact that the vertical component of the swimmer's velocity is the only one to make him reach the end of the river, and the second one from the fact that both the current and the horizontal component of the swimmer's ... 1 How Triangle Law**and **Parallelogram law of addition of Vectors are different?Ain't they. They are not different. They are the same thing. There is only one real law which is the head to tail rule. When adding any number vectors put head to tail, head to tail, head to tail... until all the vectors are used up and then draw a line from the tail to the ... 4 The vector product of a vector \vec{a} with itself is alwals zero: \vec{a} \times \vec{a} = 0 For two smooth vector-valued functions \vec{a},\vec{b} \colon \mathbb{R} \to \mathbb{R}^3 the product rule holds:$$ \frac{d}{dt} (\vec{a} \times \vec{b}) = \frac{d}{dt} \vec{a} \times \vec{b} + \vec{a} \times \frac{d}{dt} \vec{b} You can see this for ... 7 There is a identity for the derivative of the cross-product of two vector functions \mathbf A(t) and \mathbf B(t); \begin{align} \frac{d}{dt} (\mathbf A \times \mathbf B) = \frac{d\mathbf A}{dt}\times \mathbf B + \mathbf A\times \frac{d\mathbf B}{dt} \end{align} Using this rule with the computation you're considering, we obtain \begin{align} ... 0 Velocity is a vector, meaning it's direction must be taken into account. Acceleration is the change of velocity over time. On the crest the direction of the velocity changes continously, so there's an acceleration. In your hand drawing the sled passes over a horizontal terrain. Velocity does not change direction so there is no acceleration on the flat. 6 Because direction cosines are, unlike sines and tans, even functions of the angle which makes the sign of the angle irrelevant and that's a good thing. More importantly, the direction cosines of a unit vector \vec v end up being the coordinates v_x,v_y,v_z, respectively, so the direction cosines obey\cos^2 a+\cos^2 b+\cos^2 c = 1$$which is nice. ... 0 You are right that the difference between a crest and a trough is qualitatively just a mirroring. However, there is another subtle difference between the two cases. The skier is only on "level" ground for a single point along the path, while the sled on the hill is on level ground for an extended region. That is, your paths are not truly mirror images of one ... 7 In this equation F_N is the magnitude of the normal force, and W is the magnitude of the weight. The forces are in opposite directions, yes, but their magnitudes are equal. It would be correct to write this:$$\vec{F}_N = -\vec{W}$$because \vec{F}_N refers to the full force vector, including its direction, not just the magnitude. (And similarly for ... 0 If at point A with position \vec{r}_A the sum of forces and moments is \vec{F} and \vec{M}_A then the force line of action has direction$$\vec{e} = \frac{\vec{F}}{|\vec{F}|}$$and position closest to A as$$ \vec{r} = \vec{r}_A + \frac{ \vec{F} \times \vec{M}_A}{|\vec{F}|^2}$$where × is the vector cross product. This comes from the net moment ... 2 The magnitude of acceleration is simply a measurement of change in speed per unit time. As an example, say you are in a car starting from rest and you begin to speed up. Say that you reach a speed of 20 {m \over s} in 2 seconds. This means the magnitude of your acceleration is:$$ a = {20 {m \over s} \over 2s} = 10 {m \over s^2} That is, your speed ...

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Your question is kind of vague but I will try to respond. Acceleration is defined as the time rate of change of velocity. Since velocity has both magnitude and direction, so does acceleration. In other words, acceleration is a vector. The length of the vector is its magnitude. Its direction is the direction of the vector. So the magnitude of ...

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One dimensional motion The motion of an object is called one dimensional, if only one of the three co-ordinates required to specify the position of the object in space changes w.r.t time. In such a motion, the object moves along a straight line. For example, motion of a train along a straight railway track, a man walking on a level and narrow road, an ...

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Notice that in more than one dimension (two, three, and even higher!) the boldface vector notation $\mathbf x(t)$ is really just shorthand for a tuple of component functions. Explicitly, in dimension $d$, one has \begin{align} \mathbf x(t) = \begin{pmatrix} x^1(t) \\ \vdots \\ x^d(t) \\ ...

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In terms of velocitiy in classical mechanics the only difference between the dimensions is what kind of objects $v$, $u$ and $a$ are. While $t$ should always be a real number $v$,$u$ and $a$ should be vectors in $\mathbb{R}^n$ where $n$ represents the dimension you are talking about. If this does not help, please clarify.

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