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ankoles [38]
2 years ago
11

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

icle's speed at t=5.0s ? What is the particle's direction of motion, measured as an angle from the x-axis, at t=5.0s ?
Physics
1 answer:
mestny [16]2 years ago
4 0

Differentiate the components of position to get the corresponding components of velocity :

v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t

v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}

The speed at this time is the magnitude of the velocity :

\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}

The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}

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abruzzese [7]

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Perpendicular to the surface is the "normal" to the surface. So the angle of incidence (angle between the laser and the normal) is zero, and the law of refraction (just like the law of reflection) predicts an angle of zero between the normal and the refracted (or the reflected) beam.

Moral of the story:  If you want your laser to keep going in the same direction after it enters the water, or to bounce back in the same direction it came from when it hits the mirror, then shoot it <em>straight on</em> to the surface, perpendicular to it.

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2 years ago
Perpetual motion machines have fascinated people especially inventors for hundreds of years before the laws of thermodynamics be
Cerrena [4.2K]

Given the fact that energy conversion is not entirely efficient, it is impossible to produce a perpetual motion machine.

<h3>What is a perpetual motion machine?</h3>

The perpetual motion machine in one that is able to work continuously without stopping. This would mean that the efficiency of this machine must that the machine is 100% efficient which violates the second law of thermodynamics.

Thus, given the fact that energy conversion is not entirely efficient and energy looses cause machines not function effectively, it is impossible to produce a perpetual motion machine.

Learn kore about a perpetual motion machine:brainly.com/question/13001849

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5 0
2 years ago
A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.
Arturiano [62]

Answer:

a) \alpha=0.5117\ rad.s^{-2}

b) \theta=8.4\ rad

Explanation:

Given:

  • initial rotational speed of phonograph, \omega_i=0\ rad.s^{-1}
  • final rotational speed of phonograph, N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}
  • time taken for the acceleration, t=5.73\ s

a)

Now angular acceleration:

\alpha=\frac{\omega_f-\omega_i}{t}

\alpha=\frac{2.932-0}{5.73}

\alpha=0.5117\ rad.s^{-2}

b)

Using eq. of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2

\theta=8.4\ rad

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3 years ago
you throw a ball vertically so it leaves the ground withe velocity of 3.71m/s. what is its acceleration at this point
tino4ka555 [31]

No matter what direction you throw it, or with what speed, its acceleration is immediately 9.8 m/s^2 downward as soon as you release it from your hand, and it doesn't change until the ball hits something.

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3 years ago
A 0.50-kg red cart is moving rightward with a speed of 40 cm/s when it collides
lukranit [14]

The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.

The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.

The change in momentum of the system of the carts is 0.

<h3>Initial momentum of the carts before collision</h3>

The momentum of the carts before the collision is calculated as follows;

P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s

P(blue) = 1.5 x 0 = 0

<h3>Momentum of the carts after collision</h3>

The momentum of the carts after the collision is calculated as follows;

P(red) = 0.5 x 0.1 = 0.05 kgm/s

P(blue) = 1.5 0.1 = 0.15 kgm/s

<h3>Change in momentum of the carts</h3>

\Delta P = P_f - P_i

ΔP = (0.05 + 0.15) - (0.2)

ΔP = 0

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