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zepelin [54]
3 years ago
9

A uniform plank of length 5.0 m and weight 225 n rests horizontally on two supports, with 1.1 m of the plank hanging over the ri

ght support (see the drawing). to what distance x can a person who weighs 450 n walk on the overhanging part of the plank before it just begins to tip
Physics
1 answer:
likoan [24]3 years ago
6 0
<span>0.7 meters from the support, or 0.4 meters from the end of the board. Another way of describing the problem is to have the 450 N person move so that the center of mass of the plank and person system is on the second support. So let's first determine how far from that support the center of mass of the plank is. We can model the plank as a point mass source that's 2.5 meters from either end. So that point mass is 2.5-1.1 = 1.4 meters from the pivot. So let's multiply that radius by the mass, getting: 1.4 m * 225 N = 315 Nm Now the person will be at just the tipping point when their mass multiplied by the distance equal 315. So let's do the division. 315 Nm / 450 N = 0.7 m So the person will be at the tipping point when he's 0.7 meter from the support, or 1.1 - 0.7 = 0.4 meters from the end of the board.</span>
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Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

\theta = Angle = 20°

y = -20 cm

Velocity components

u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s

u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s

Acceleration components

a_x=0

a_y=-9.81\ m/s^2

y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0

t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629

Time taken is 0.26088 seconds

x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m

The distance the beetle travels on the ground is 0.3677181864 m

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An electron enters a region of space containing a uniform 0.0000193-T magnetic field. Its speed is 121 m/s and it enters perpend
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Answer

Given,

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speed, v = 121 m/s

mass of electron, m = 9.11 x 10⁻³¹ Kg

charge of electron, q = 1.6 x 10⁻¹⁹ C

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3 years ago
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