2 years ago

could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad

Physics

1 answer:

dexar [7]2 years ago

4 0

Answer:

f = 4000 / 30 sec = 133.3 vibrations/sec

P = 1 / f = .0075 sec period of 1 vibration

You might be interested in

What is the velocity of an object that has a momentum of 4,000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.

julia-pushkina [17]

<h2>Answer: 34.78 m/s</h2>

Explanation:

The momentum $p$ is given by the following equation:

$p=m.V$ (1)

Where:

$m$ is the mass of the object

$V$ is the velocity of the object

Finding the velocity from (1):

$V=\frac{p}{m}$ (2)

$V=\frac{4000kg.m/s}{115kg}$

<u>Finally:</u>

$V=34.78m/s$ >>>This is the velocity of the object

4 0

3 years ago

Read 2 more answers

In a line graph the dependent variable is plotted on the

gladu [14]

Answer:

X axis

Explanation:

I hope it helps :)

3 0

2 years ago

Why is the level of glycerin initially decreases when heated ?

Mila [183]

Because glucerin is hotter so it decreases faster

4 0

2 years ago

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

4 0

3 years ago

Two objects, with different sizes, masses, and temperatures, are placed in thermal contact. in which direction does the energy t

s344n2d4d5 [400]

<span>(c) energy travels from the object at higher temperature
to the object at lower temperature.

Size and mass have no effect.</span>

3 0

3 years ago