If we can measure the period of a star's wobble caused by an orbiting planet, we know the ______

A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th

solniwko [45]

Answer:

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

$a_t = R\alpha$

2) Centripetal acceleration

$a_c = \omega^2 R$

here we know that

$\omega = \alpha t$

$a_c = (\alpha t)^2 R$

now we know that net linear acceleration is given as

$a = \sqrt{a_c^2 + a_t^2}$

so we have

$a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}$

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

4 0

3 years ago

HELPPPPP !!!!

Nataly [62]

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

4 0

3 years ago

What is the SI unit for kinetic energy

trapecia [35]

The easiest way to build a unit for energy is to remember that
'work' is energy, and
   Work = (force) x (distance).

So energy is    (unit of force) x (unit of distance)

   [Energy] =   (Newton) (meter) .

'Newton' itself is a combination of base units, so
energy is really
   (kilogram-meter/sec²) (meter)

   = kilogram-meter² / sec² .

That unit is so complicated that it's been given a special,
shorter name:
   Joule .

It doesn't matter what kind of energy you're talking about.
Kinetic, potential, nuclear, electromagnetic, food, chemical,
muscle, wind, solar, steam ... they all boil down to Joules.

And if you generate, use, transfer, or consume 1 Joule of
energy every second, then we say that the 'power' is '1 watt'.

3 0

3 years ago

Read 2 more answers

Please help me out if you want brainliest !!!!! ASAP easy

Vinvika [58]

Answer:

Road A = 8 meters Wet

Road B = 2 meters Muddy

Road C = 12 meters Dry

Explanation:

8 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 <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

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

4 0

3 years ago

If we can measure the period of a star's wobble caused by an orbiting planet, we know the _______.