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guapka [62]
2 years ago
10

The default unit of rotational speed is 'radians per second' or 'rad/s'. If a someone sits at the edge of a 5m radius carousel a

nd has tangential speed of 2 m/s, what is the rotational speed in rad/s?
A. 2.5
B. 3
C. 7
D. 10
Physics
1 answer:
Blizzard [7]2 years ago
7 0

The rotational speed of the person is 0.4 rad/s.

<h3>Rotational speed (rad/s)</h3>

The rotational speed of the person in radian per second is calculated as follows;

v = ωr

where;

  • v is linear speed in m/s
  • r is radius in meters
  • ω is speed in rad/s

ω = v/r

ω = 2/5

ω = 0.4 rad/s

Thus, the rotational speed of the person is 0.4 rad/s.

Learn more about rotational speed here: brainly.com/question/6860269

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Degger [83]

Answer:

Angle of refraction

Explanation:

The incident ray is the ray before it reaches the surface.

The refracted ray is the ray after it reaches the surface.

n₁ is called the index of incidence.

n₂ is called the index of refraction.

θ₁ is called the angle of incidence.

θ₂ is called the angle of refraction.

They are related by Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

5 0
3 years ago
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What is 62.125 kg rounded to two significant figures?
Mamont248 [21]

<em> </em><em> </em><em>62.125</em>

<em>=</em><em> </em><em>62.12</em>

<em>THANK</em><em> </em><em>you</em><em> </em>

6 0
3 years ago
g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period
NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

   $e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

   $\frac{R}{2L}T= \ln \frac{5}{3.8}$

  $\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

 $\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

                     $=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

                     $=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

              $\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

  $Q=\sqrt{131.166}$

      = 11.45

4 0
3 years ago
The figure (Figure 1) shows the velocity of a solar-powered motorhome (RV) as a function of time. The driver accelerates from a
SpyIntel [72]

Explanation :

It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.

We know that acceleration is defined as the rate of change of velocity.

a=\dfrac{dv}{dt}

Where

dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s

dt is the change in time, dt = 40 s - 30 s = 10 s

So, a=\dfrac{-60\ m/s}{10\ s}

a = -6\ m/s^2

From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. -6\ m/s^2.

6 0
3 years ago
There are 2 concentric cylinders. These cylinders are very long with length L. The inner cylinder has a radius R1 and is a solid
OLga [1]
To get the charge along the inner cylinder, we use Gauss Law
E = d R1/2εo
For the outer cylinder the charge can be calculated using
E = d R2^2/2εoR1
where d is the charge density
Use these two equations to get the charge in between the cylinders and the capacitance between them.
5 0
3 years ago
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