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Vlad [161]
3 years ago
13

A wheel of radius 14 inches is rotating 0.5 rad/s. what is the linear speed v, the angular speed in rpm, and the angular speed i

n deg/s?
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

First, let us find for the linear speed v. The formula for this is:

v = r w

where r is radius and w is radial velocity, therefore the linear speed is:

v = 14 inches (0.5 rad / s)

v = 7 inches / s

 

Now calculating for angular speed in rpm or revolutions per minute:

angular speed in rpm = (0.5 rad / s) (1 rev / 2π rad) (60 s / min)

<span> angular speed in rpm = (15/π) rpm = 4.77 rpm</span>

 

Finally calculating for angular speed in deg per second:

angular speed in rpm = (0.5 rad / s) (360° / 2π rad)

<span> angular speed in rpm = (90/π) deg/s = 28.65 deg/s</span>

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At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta
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Some of our appliances have a magnet in it.<br><br>yes or no​
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Answer:

yes !!

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Magnets can be found in the simplest or most complex devices you use every day. From home appliances such as the refrigerator, microwave oven and electric fan, to your company's office equipment such as computers and printers. All these devices use magnets.

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In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin
Natasha2012 [34]

Answer:

y = 0.0233 m

Explanation:

In a Young's Double Slit Experiment the distance between two consecutive bright fringes is given by the formula:

Δx = λL/d

where,

Δx = distance between fringes

λ = wavelength of light

L = Distance between screen and slits

d = Slit Separation

Now, for initial case:

λ = 425 nm = 4.25  x 10⁻⁷ m

y = 2Δx = 0.0177 m => Δx = 8.85 x 10⁻³ m

Therefore,

8.85 x 10⁻³ m = (4.25 x 10⁻⁷ m)L/d

L/d = (8.85 x 10⁻³ m)/(4.25 x 10⁻⁷ m)

L/d = 2.08 x 10⁴

using this for λ = 560 nm = 5.6 x 10⁻⁷ m:

Δx = (5.6 x 10⁻⁷ m)(2.08 x 10⁴)

Δx = 0.0116 m

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y = 2Δx

y = (2)(0.0116 m)

<u>y = 0.0233 m</u>

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3 years ago
lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

5 0
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