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2 years ago

By what factor should the length of a simple pendulum be changed in order to triple the period of vibration?

Physics

1 answer:

kenny6666 [7]2 years ago

4 0

Answer:

increased by 9 times

Explanation:

As the period of a simple pendulum is defined as the following formula

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

where L is the length of the pendulum and g is the constant gravitational acceleration. Since L is under a square root, if you want to triple the period of vibration, L must be increased by 9 times so that $\sqrt{9} = 3$

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What is difference between kilowatt and kilowatt hour?

Scrat [10]

Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate. The other is a quantity, after maintaining a rate for some time.

-- 'Joule' is a unit of energy. It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.

-- 'Watt' is a <u><em>rate</em></u> of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your <em>power</em> is 40.7 watts.

-- 'Watt-second' is the amount of energy used in one second,
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-- 'Watt-hour' is the amount of energy used in one hour,
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-- 'Kilowatt - second' is the amount of energy used in one second,
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Depending on where you live, 3,600,000 joules of energy bought
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6 0

2 years ago

Read 2 more answers

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

Where;

$G$ is the Gravitational Constant and its value is $6.674(10^{-11})\frac{m^{3}}{kgs^{2}}$

$M=1.9(10^{27})kg$ is the mass of Jupiter

$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}$

$T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}$

$T=\sqrt{2.339(10^{10})s^{2}}$

Then:

Which is the same as:

Therefore, the answer is:

The orbital period of Io is 42.482 h

7 0

3 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

Type the correct answer in the box. Spell the word correctly.

serious [3.7K]

Answer:

The Lambda-CDM model contains a cosmological constant, denoted by a lambda (λ), which is associated with dark energy and <u>cold dark matter</u>.

^Also works for Plato users.

3 0

3 years ago

Read 2 more answers

What is a likely consequence of continued human population growth

quester [9]

Answer:

Rapid human population growth has a number of consequences. Population grows fastest in the world's poorest countries. High fertility rates have historically had a strong relation with poverty, and high childhood mortality rates

Explanation:

i got it from goo.gle but i hope that helps:)

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2 years ago