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3 years ago

When the velocity of an object does not change, what happens to its acceleration?

Physics

1 answer:

Ratling [72]3 years ago

4 0

Answer:

Acceleration is the rate of change of velocity.

When the velocity of an object does not change, i.e., when the velocity is constant, the acceleration is <em>zero</em>.

Mathematically (differential form),

$Velocity = k, \: where \: k \: is \: a \: constant \\ \frac{d(Velocity)}{dt} = 0 \\\boxed{ Acceleration = 0}✓$

<h3><u>Zero</u> is the right answer.</h3>

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3 years ago

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A 0.0208 m diameter coin rolls up a 18.0◦ inclined plane. The coin starts with an initial angular speed of 56.0 rad/s and rolls

anastassius [24]

Answer:

h = 0.0259 m

Explanation:

given,

diameter of the cone = 0.0208 m

radius,r = 0.0104 m

angle of inclination,θ = 18°

initial angular velocity, ω_i = 56 rad/s

final angular velocity ,ω_f = 0 rad/s

height, h = ?

Rotational kinetic energy

$KE_r = \dfrac{1}{2}I\omega^2$

Moment of inertia of coin

$I = \dfrac{1}{2}MR^2$

so,

$KE_r = \dfrac{1}{4}MR^2\omega^2$

Transnational Kinetic energy

$KE_t = \dfrac{1}{2}Mv^2$

v = r ω

$KE_t = \dfrac{1}{2}MR^2\omega^2$

now,

using conservation energy

Kinetic energy of the coin is converted into the potential energy

KE_r + KE_t = PE

$\dfrac{1}{4}MR^2\omega^2 + \dfrac{1}{2}MR^2\omega^2 = Mgh$

$\dfrac{3}{4}R^2\omega^2=gh$

$\dfrac{3}{4}\times 0.0104^2\times 56^2=9.8\times h$

h = 0.0259 m

Vertical height gain by the coin is equal to 0.0259 m

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4 years ago

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Mrac [35]

Answer:

the 2nd one

Explanation:

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