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densk [106]
3 years ago
11

A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calcu

late the angular acceleration​
Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:

2 rad/s²

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 1 Kg

Radius (r) = 0.2 m

Angular speed (w) = 20 rad/s

Time (t) = 10 s

Angular acceleration (a) =?

Angular acceleration is defined as:

Angular acceleration (a) =Angular speed (w) / time

a = w/t

With the above formula, we can obtain the angular acceleration of the sphere as follow:

Angular speed (w) = 20 rad/s

Time (t) = 10 s

Angular acceleration (a) =?

a = w/t

a = 20 / 10

a = 2 rad/s²

Thus, the angular acceleration of the sphere is 2 rad/s²

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Some of our appliances have a magnet in it.<br><br>yes or no​
Montano1993 [528]

Answer:

yes !!

Explanation:

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A discharge lamp rated at 25 W (1 W = 1 J/s) emits yellow light of wavelength 580 nm. How many photons of yellow light does the
Scrat [10]

Answer:

the number of photons of yellow light does the lamp generate in 1.0 s is 7 x 10^{19}

Explanation:

given information:

power, P = 25 W

wavelength. λ - 580 nm = 5.80 x 10^{-7} m

time, t = 1 s

to calculate the number of photon(N), we use the following equation

N = λPt/hc

where

λ = wavelength (m)

P = power (W)

t = time interval (s)

h = Planck's constant (6.23 x 10^{-34} Js)

c = light's velocity (3 x 10^{8} m/s^{2})

So,

N = λPt/hc

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3 0
2 years ago
Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r
ella [17]

Answer:

\frac{dR(t)}{dt}=0.06\Omega

Explanation:

Since R(t)=\frac{V}{I(t)}, we calculate the resistance rate by deriving this formula with respect to time:

\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})

Deriving what is left (remember that (\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)):

\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}

So we have:

\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega

8 0
3 years ago
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