Ask question

Ask question

All categories

3 years ago

5

A simple pendulum consists of a 2 kg bob attached to a 1.5 m long string. How much time (in s) is required for this pendulum to

complete 2.5 oscillations?

1 answer:

Charra [1.4K]3 years ago

3 0

Answer:

6.15 s

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

For the pendulum in this problem,

L = 1.5 m (length)

$g=9.8 m/s^2$ (acceleration due to gravity on Earth)

Therefore, its period is

$T=2\pi \sqrt{\frac{1.5}{9.8}}=2.46 s$

And therefore, the time taken for the pendulum to complete 2.5 oscillations is equal to 2.5 times the period:

$t=2.5T=(2.5)(2.46)=6.15 s$

You might be interested in

A book falls off a desk. The book has a mass of 1.2 kg and was at a height of 0.9 m. How much mechanical energy does the book ha

Valentin [98]

The book's potential energy will be converted entirely to kinetic energy at that point in the fall, since we can neglect losses due to air resistance. The potential energy is given by PE=mgh, where m=1.2kg, h=0.9m, and g=9.81m/s^2. This is the acceleration due to gravity.
This gives:
PE=1.2*0.9*9.81=10.59J
This is also the kinetic (mechanical) energy right before the landing.

3 0

3 years ago

Read 2 more answers

25 equals 1/2×2× V squared

solong [7]

Answer:

v=5

Explanation:

1/2 x 2 is 1

1*5^2 is 25

5^2 is 25

6 0

4 years ago

A conducting strip of width 1.5 mm is in a magnetic field. As a result, there is a potential difference of 4.3 mV across the wid

OLga [1]

To solve this problem we will apply the concepts related to the Voltage depending on the Electric field and the distance where it is applied. The relationship of this relationship is given in the following form:

$V= Ed \rightarrow E = \frac{V}{d}$

Here,

E = Electric Field

V = Voltage

d = Distance

Our values are given as,

$d = 1.5mm = 1.5*10^{-3} m$

$V = 4.3 mV$

$V = 4.3*10^{-3}V$

Replacing at the first equation we have,

$E = \frac{4.3*10^{-3}}{1.5*10^{-3}}$

$E = 2.866V/m$

Therefore the electric field in the strip is 2.866V/m

6 0

3 years ago

The planet Mars has a mass of 6.1 × 1023 kg and radius of 3.4 × 106 m. What is the acceleration of an object in free fall near t

Oksanka [162]

The acceleration due to gravity of Mars is $3.5\ m / s^{2}$

<u>Explanation:</u>

As per universal law of gravity, the gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them. But in the present case, the gravity need to be determined between Mars and the object on Mars. Since the mass of Mars is greater than the mass of any object. Thus,

$\text {Gravitational force of planet}=\frac{G M m}{R^{2}}$

Here, G is the gravitational constant, R is the radius of Mars and M, m is the mass of Mars and the object respectively..

Also, according to Newton’s second law of motion, the acceleration of any object will be equal to the ratio of force exerted on it to the mass of the object.

So in order to determine the acceleration due to gravity of Mars, divide the gravitational force of Mars by mass of object on the surface of Mars.

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{\text {Gravitational force of Mars}}{m \text { of object }}$

$Acceleration\ due\ to\ gravity\ of\ mars =\frac{G M m}{R^{2} \times m}=\frac{G M}{R^{2}}$

$\text { Acceleration due to gravity of mars }=\frac{6.67259 \times 10^{-11} \times 6.1 \times 10^{23}}{3.4 \times 3.4 \times 10^{12}}=\frac{40.703 \times 10^{12}}{11.56 \times 10^{12}}$

$\text { Acceleration }=3.5\ \mathrm{m} / \mathrm{s}^{2}$

3 0

3 years ago

A harmonic wave on a string is described by

xxMikexx [17]

$\huge{\bold{\orange{\underline{ Solution }}}}$

<h3><u>Given </u><u>:</u><u>-</u></h3>

A harmonic wave on a string is described by

$\sf{ Y( x, t) = 0.1 sin(300t + 0.01x + π/3)}$

x is in cm and t is in seconds

<h3><u>Answer </u><u>1</u><u> </u><u>:</u><u>-</u></h3>

<u>Equation </u><u>for </u><u>travelling </u><u>wave </u><u>:</u><u>-</u>

$\sf{ Y( x, t) = Asin(ωt + kx + Φ)...eq(1)}$

<u>Equation</u><u> </u><u>for </u><u>stationary </u><u>wave </u><u>:</u><u>-</u>

$\sf{ Y( x, t) = Acos(ωt - kx )...eq(2)}$

<u>Given </u><u>equation </u><u>for </u><u>wave </u><u>:</u><u>-</u>

$\sf{ Y( x, t) = 0.1 \:sin(300t + 0.01x + π/3)...eq(3)}$

<u>On </u><u>comparing </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>,</u><u> </u><u>(</u><u>2</u><u>)</u><u> </u><u>and </u><u>(</u><u>3</u><u>)</u>

We can conclude that, Given wave represent travelling wave.

<h3><u>Answer </u><u>2</u><u> </u><u>:</u><u>-</u></h3>

From solution 1 , We can say that,

$\sf{ Y( x, t) = 0.1 \: sin(300t + 0.01x + π/3).}$

It is travelling from right to left direction

Hence, The direction of its propagation is right to left that is towards +x direction.

<h3><u>Answer </u><u>3</u><u> </u><u>:</u><u>-</u></h3>

Here, We have to find the wave period

<u>We </u><u>know </u><u>that</u><u>, </u>

Wave period = wavelength / velocity

<u>Wave </u><u>equation</u><u> </u><u>:</u><u>-</u>

$\sf{ Y( x, t) = 0.1 \:sin(300t + 0.01x + π/3).}$

ω = 300rad/s
k = 0.01

<u>We </u><u>know </u><u>that</u><u>, </u>

$\sf{v =}{\sf{\dfrac{ ω}{2π}}}{\sf{\: and\:}}{\sf{ λ =}}{\sf{\dfrac{ 2π}{k}}}$

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ wave\: period =}{\sf{\dfrac{ 2π/k}{ω/2π }}}$

$\sf{ wave \:period = }{\sf{\dfrac{k}{ω}}}$

$\sf{ wave\: period =}{\sf{\dfrac{ 0.01}{300}}}$

$\sf{ wave\: period = 0.000033\: s}$

<h3><u>Answer </u><u>4</u><u> </u><u>:</u><u>-</u></h3>

The wavelength of given wave

$\bold{ λ = }{\bold{\dfrac{2π}{k}}}$

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ λ = }{\sf{\dfrac{2 × 3.14 }{0.01}}}$

$\sf{ λ = }{\sf{\dfrac{6.28}{0.01}}}$

$\sf{ λ = 628 \: cm }$

<h3><u>Answer </u><u>5</u><u> </u><u>:</u><u>-</u></h3>

We have wave equation

$\sf{ Y( x, t) = 0.1 sin(300t + 0.01x + π/3).}$

<u>Travelling </u><u>wave </u><u>equation </u><u>:</u><u>-</u>

$\sf{ Y( x, t) = A\:sin(ωt + kx + Φ)...eq(1)}$

<u>Therefore</u><u>, </u>

Amplitude of the wave particle

$\sf{ A = 0.1 \: cm}$

Hence, The amplitude of the particle is 0.1 cm

8 0

2 years ago