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iVinArrow [24]
3 years ago
11

Observations show that the dragonflies complete one revolution every 0.017 s during this "spin dry" maneuver. What is their angu

lar speed in radians per second?
Physics
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

\omega=369.6\frac{rad}{s}

Explanation:

The time taken to complete one revolution is the period and is defined as:

T=\frac{1}{f}(1)

The angular speed is given by:

\omega=2\pi f\\f=\frac{\omega}{2\pi}(2)

Replacing (2) in (1) and solving for \omega:

T=\frac{1}{\frac{\omega}{2\pi}}\\\omega=\frac{2\pi}{T}\\\omega=\frac{2\pi}{0.017s}\\\omega=369.6\frac{rad}{s}

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When the temperature increases, the intermolecular forces between the molecules of a liquid become weaker, and some bonds break easily. Thus as temperature increases, the surface tension of a liquid decreases.
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A spring does 5.0 J of work on a 0.10-kg ball bearing in a pinball machine. The ball's
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Answer:

10m/s

Explanation:

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3 years ago
A 3 x 10^-6 C charge is 5m away from a -2x 10^-6 C charge A. Attractive because one is positive the other one is negative B. Det
Marizza181 [45]

Answer:

0.0021576N

Explanation:

F=(k)(q1q2/r^2)

F=(8.99×10^9)(3×10^-6)(2×10^-6)/(5^2)

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2 years ago
A rock is thrown downward from an unknown height above the ground with an initial speed of 6.1 m/s. It strikes the ground 1.7 s
insens350 [35]

Answer:

24.531 m

Explanation:

t = Time taken = 1.7 s

u = Initial velocity = 6.1 m/s

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=6.1\times 1.7+\dfrac{1}{2}\times 9.8\times 1.7^2\\\Rightarrow s=24.531\ m

The initial height of the rock above the ground is 24.531 m

7 0
2 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

\Phi = BA Cos \theta

Here,

\theta = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

\epsilon= -N \frac{d\Phi }{dt}

\epsilon = -N \frac{(BAcos\theta)}{dt}

\epsilon = -NAcos\theta \frac{dB}{dt}

\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)

\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )

At time t = 5.71s,  Induced emf is,

\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

\epsilon = 10.9V

Therefore the magnitude of the induced emf is 10.9V

4 0
3 years ago
Read 2 more answers
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