____ 65. a green laser pointer emits 532 nm light. what is the frequency of this radiation

A 1000 kg race car rounds a curve with a radius of 50 m.

Kruka [31]

Answer:

5644556677888777766554433

5 0

2 years ago

Compare the momentum of a 6,300-kg elephant walking 0.11 m/s and a 50-kg dolphin swimming 10.4 m/s. your answer

bogdanovich [222]

First sum applied the Newton's second law motion: F = ma Force = mass* acceleration This motion define force as the product of mass times Acceleration (vs.Velocity). Since acceleration is the change in velocity divided by time, force=(mass*velocity)/time such that, (mass*velocity)/time=momentum/time Therefore we get mass*velocity=momentum Momentum=mass*velocity Elephant mass=6300 kg; velocity=0.11 m/s Momentum=6300*0.11 P=693 kg (m/s) Dolphin mass=50 kg; velocity=10.4 m/s Momentum=50*10.4 P=520 kg (m/s) The elephant has more momentum(P) because it is large.

5 0

3 years ago

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

mestny [16]

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 t = 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 x-axis, such that

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