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As shown in the figure, the rotating water in the bucket will become concave, and the faster the rotation, the more concave the water surface is. This shows that at a certain height H, the closer the water is to the center of rotation, the lower the pressure is. And the faster the water rotates, the lower the pressure of the water in the center of rotation. ...


3

I'm not sure what you mean by "where does that energy come from?" The Magnus force is a force not an energy. As the force is at right angles to the motion, the Magnus force does no work, so no energy need come from anywhere.


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As shown in the figure, the water surface in a rotating bucket will eventually become concave. The faster the rotating speed, the more concave the water surface is. The bucket in the left picture does not rotate very fast, so the water surface is not concave very much. The bucket in the right picture rotates very fast, so the water surface is concave very ...


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About centrifugal effect Let's start at the Equator. Your distance to the Earth axis of rotation is the Earth radius (at the Equator), and your angular velocity is one revolution per day. The required centripetal acceleration for that specific circumnavigating motion is about 0.034 m/s^2 That required centripetal acceleration goes at the expense of the ...


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I'll take the lowest values as this is a question about effectiveness, so let's assume no air friction and that the bow doesn't dissipate energy in any other form than pure kinetic energy. The force felt was 70kg so converting that to newtons gives us 686.70 newtons dragging the 0.94m long arrow back averaging 343.35 (correction by alephzero) thus giving us ...


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In the general case, we can write the drag force in vector form $$\vec F_D=\frac {1}{2}\rho A C_D(Re)|\vec v_0-\vec v|(\vec v_0-\vec v)$$ here $\vec v_0$ is wind speed, $\vec v$ is body speed, $Re$ is Reynolds number. In the particular case of $\vec v_0=0$, we have $$\vec F_D=-\frac {1}{2}\rho A C_D(Re)|\vec v|\vec v$$ The projections of the force on the ...


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Drag acts in the opposite direction of velocity. Therefore, if $$v_x = V \cdot \cos\left(\frac{\pi}{6}\right)$$ $$v_y= V \cdot \sin\left(\frac{\pi}{6}\right)$$ Then $$D_x = -D \cdot \cos\left(\frac{\pi}{6}\right)$$ $$D_y =-D \cdot \sin\left(\frac{\pi}{6}\right)$$ This is because, in general, if for vectors $\mathbf a$ and $\mathbf b$ the following is true ...


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Intrigued by @Bill Watts answer I decided to do this answer step by step and found out that Bill was indeed right. But I also observed that we all were also right but the thing was that we weren't actually calculating the values on pen and paper and hence believed that multiplying the initial velocity by 10 would be wrong. Anyway here is the solution


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Let us start with basic what question actually want since man has mass of 100kg (including gun mass) ,the question is all about momentum conservation since no impulsive force is present momentum relationship. F Δt=0 so initial momentum =final momentum so lets start when bullet one is fired initial momentum was zero so the final momentum should also.equate ...


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We can do the same for each subsequent shot, because the momentum change with each subsequent shot is the same as the momentum change with the first shot. Look at the impact = momentum relationship. $F\ \Delta t =m\ \Delta v$ Each shot produces the same $F\ \Delta t$ and therefore produces the same momentum change in the man and his muzzle. Another way ...


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Imagine only one bullet is being fired. If (after the firing) the velocity of this bullet is $v_b$ and the velocity of the system (man + remaining bullets) is $v_s$, then we apply momentum conservation as follows: $m_Sv_s + m_bv_b = 0$ or $\frac{v_s}{v_b} = -\frac{m_b}{m_s} $ Therefore, after firing one shot, the velocities of the man and the bullet, will ...


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You are almost there. The two equations you provided are missing the $\theta$ term. It should be written: $ x(t)=x_0+\left(v_0\cos\theta \right)t\\ y(t)=y_0+\left(v_0\sin\theta \right)t-\frac{1}{2}gt^2 $ You can also write the equations for the velocity, whose components are: $ v_x(t)=v_0\cos\theta \\ v_y(t)=v_0\sin\theta - gt $ Now, "the projectile hits ...


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Considering the movement origin at the top of $h_1 \le h_2$ and calling $v_1^2 = v_x^2+v_y^2$ the required initial velocity at this point, we have $$ \frac{v_y}{v_x}(x_2-x_1)-\frac 12 \frac{g}{v_x^2}(x_2-x_1)^2 = h_2-h_1 $$ then $$ v_y = v_x\frac{h_2-h_1}{x_2-x_1}+\frac 12\frac{g}{v_x}(x_2-x_1) $$ but calling $v_0$ the ground initial velocity we have $$ ...


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If the initial velocity is $v_0$, then we have $v(s) = \sqrt{v_0^2-2gs}$. At $s_0 = \frac{v_0^2}{2g}$ we have $v(s_0) = 0$ but the function $v(s) = \sqrt{v_0^2-2gs}$ is not differentiable at $s_0$. Therefore the relation $$a(s) = v(s)\cdot \frac{dv(s)}{ds}$$ is actually not true at $s_0$, because $\frac{dv(s)}{ds}\Big|_{s=s_0}$ does not exist.


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There's nothing wrong with $a=v \frac{dv}{ds}$. At the highest point, $v=0$, which makes $dv/ds$ undefined. If you take the appropriate limit, you get $a=-g$ again. Let me elaborate. Start with the standard uniformly accelerated linear motion equations: $$v=v_0 + at$$ $$s = s_0 + v_0 t + \frac{1}{2}at^2$$ Your formula for $a$ then gives $$ a = v \frac{dv}...


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The clearest way to tackle this question is to remember that the momentum is always zero. If the gun is floating in mid-air, then the total momentum of "gun plus bullet" is zero before the gun is fired and zero after the gun is fired. So if you know the forward velocity of the bullet, you can multiply by the mass of the bullet to get the forward momentum ...


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