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Marizza181 [45]
3 years ago
15

A 2kg bowling ball is 2.5 meter off the ground on a post when it falls. Just before it reaches the ground, it is traveling 7m/s.

Assuming that there is no air resistance, which statement is true
Physics
1 answer:
shutvik [7]3 years ago
7 0
<span>The mechanical energy is conserved.

I hope this helps, good luck! :)</span>
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Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.
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Answer:

\tau=(-32k)\ N-m

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Explanation:

Given that.

Force acting on the particle, F=(-6i + 2.0j)\ N

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To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

\tau=F\times r

\tau=(-6i + 2.0j)\times (4i+4j)

\tau=\begin{pmatrix}0&0&-32\end{pmatrix}

\tau=(-32k)\ N-m

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r, |r|=\sqrt{4^2+4^2}=5.65\ m

Magnitude of F, |F|=\sqrt{(-6)^2+2^2}=6.324\ m

Using dot product formula,

F{\circ}\ r=|F|.|r|\ cos\theta

cos\theta=\dfrac{F{\circ} r}{|F|.|r|}

cos\theta=\dfrac{-24+8}{6.324\times 5.65}

\theta=116.55^{\circ}

Therefore, this is the required solution.

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