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Slav-nsk [51]
1 year ago
12

For an object being launched at ground level at a tilted non-zero initial velocity, is the velocity zero anywhere? are the horiz

ontal or vertical components of the velocity zero at any point?
Physics
1 answer:
borishaifa [10]1 year ago
3 0

The correct answer No the velocity is not zero anywhere For an object being launched at ground level at a tilted non-zero initial velocity, and horizontal or vertical components of the velocity is zero at maximum height .

Because velocity is a vector (it has magnitude and direction), the overall velocity of an object can be calculated by adding the x and y components: v2 = vx2 + vy2. Meters are used to express horizontal and vertical distances (m). The horizontal velocity component (vx) describes the effect of velocity on horizontal displacement of the projectile.

The vertical velocity component (vy) describes the effect of velocity on vertical displacement of the projectile. The vertical component of a projectile's velocity is momentarily zero (vy = 0 m/s) when it reaches maximum height. The horizontal component of its velocity, however, is not zero.

Learn more about components of velocity here :-

brainly.com/question/12498991

#SPJ4

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Read 2 more answers
I shared a picture of the problem. It’s a basic Physics question and an Algebra question.
julia-pushkina [17]

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

Given the expressions;

T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}

Equating both expressions we will have;

2 \pi \sqrt{\frac{m}{k} }  = \frac{2 \pi}{\omega}

Divide both equations by 2π

\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\

Square both sides

(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}

Take the square root of both sides

\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

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3 years ago
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