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1 year ago

What is the period and frequency of the second hand on a clock?

Physics

1 answer:

Tanzania [10]1 year ago

6 0

Answer:

Explanation:For second hand, period = 1 min = 3660 s; frequency = 0.0166 Hz}\\\mbox{For minute hand, period = 1 hr = 3600 s; frequency = 2.77}\times10^{-4} Hz\\\mbox{For hour hand, period = 12 hrs = 4300 s; frequency = 2.3}\times10^{-5} Hz.

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According to Bohr each orbit an electron could travel in had a certain amount of ________ associated with it.

Lena [83]

The awnser to your problem would be a as in energy

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2 years ago

A hot-air balloon is ascending at the rate of 12 m/s and is 80 m above the ground when a package is dropped over the side. (a) H

Debora [2.8K]

Answer:

Explanation:

Given

balloon is ascending at the rate of $u=12\ m/s$

Balloon is at a height of $s=80\ m$

As the package is dropped from the balloon it must possess the same initial velocity as the balloon

using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2=12^2+2\times 9.8\times 80$

$v^2=1712$

$v=41.37\ m/s$

time taken is given by

$v=u+at$

substituting values

$41.37=-12+9.8\times t$

as we consider downward direction as positive

$t=\frac{53.37}{9.8}$

$t=5.44\ s$

therefore time taken to reach the bottom is t=5.44 s

6 0

2 years ago

Which of the following statements is not true regarding atmospheric pressure?

NISA [10]

When atmospheric pressure is higher than the absolute pressure of a gas in a container.

8 0

2 years ago

Read 2 more answers

A top of rotational inertia 4.0 kg m2 receives a torque of 2.4 nm from a physics professor. the angular acceleration of the body

Tanzania [10]

Angular acceleration is simply the ratio of the Torque over the rotation inertia, that is:

Angular acceleration = Torque / Rotational inertia

So substituting the values:

Angular acceleration = 2.4 N m / 4.0 kg m2

<span>Angular acceleration = 0.7 rad/s^2</span>

5 0

2 years ago

A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha

ipn [44]

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, $L=400\ \mu H=400\times 10^{-6}\ H$

Capacitance of parallel LCR circuit, $C=0.3\ \mu F=0.3\times 10^{-6}\ F$

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

$f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }$

$f=14528.79\ Hz$

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

4 0

2 years ago