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4 years ago

A ball is tossed up in the air. at its peak, it stops before beginning to fall. the ball at its peak has

2 answers:

3 0

The projectile (ball) reaches an instantaneous vertical speed (Vy) of zero at maximum height.

so, V(max height) = ¬Г(Vx)^2+(Vy)^2
in this case V(max height) = Vx, where Vy=0

The maximum height, Yf, can be solved using Vfy^2=Viy^2 + 2gy. At maximum height Vfy=0.

melisa1 [442]4 years ago

3 0

Answer:potential energy only?

Explanation: the ball is at the top which stores energy like potential energy

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3 years ago

The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general

Sever21 [200]

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

$O=(0,0)$

$A=(0,-b)$

$B=(h,0)$

We need to calculate the position vector of AB

$\bar{AB}=(h-0)i+(0-(-b))j$

$\bar{AB}=hi+bj$

We need to calculate the unit vector along AB

$u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}$

$u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}$

We need to calculate the force acting along the edge

$\hat{F}=F(u_{AB})$

$\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})$

We need to calculate the net moment

$\hat{M}=\hat{F}\times OA$

Put the value into the formula

$\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})$

$\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}$

Put the value into the formula

$\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}$

$\hat{M}=-81.102\ \hat{k}\ N-m$

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

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3 years ago

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2 years ago

If a permanent magnet is dropped or struck by a hammer, it may loose it's magnetism. Explain why.

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3 years ago

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It starts to melt the minerals in within the earths core.

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