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3 years ago

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Consider a rigid body that rotates but whose center of mass is at rest. a. Trueb. false: the rotational kinetic energy of the en

tire object is equivalent to the sum of the translational kinetic energy of each small piece of the object.

1 answer:

Fiesta28 [93]3 years ago

5 0

Answer:

Explanation:

True

It is correct as the total rotational energy is the sum of transnational kinetic energy of each small piece of the object.

Consider a rigid body made of small particles each of mass dm have same angular velocity ω and the radius of the rotational path is r.

The velocity of each particle is v = r ω

So, the kinetic energy of each particle is

dK = 0.5 dm x v² = 0.5 x dm x r² x ω²

The total kinetic energy

K = 0.5 x I ω²

where, I is the moment of inertia

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You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

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2 years ago

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Answer:

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Explanation:

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Varvara68 [4.7K]

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Anna11 [10]

Answer:2.72 m/s

Explanation:

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Length of inclined Plane $s=5 m$

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3 years ago