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Four mass–spring systems oscillate in simple harmonic motion. Rank the periods of oscillation for the mass–spring systems from l

liubo4ka [24]

The periods of oscillation for the mass–spring systems from largest to smallest is:

m = 4 kg , k = 2 N/m (T = 8.89 s)
m = 2 kg , k = 2 N/m (T = 6.28 s)
m = 2 kg , k = 4 N/m (T = 4.44 s)
m = 1 kg , k = 4 N/m (T = 3.14 s)

<h3>Explanation:</h3>

The period of oscillation in a simple harmonic motion is defined as the following formulation:

$T=2\pi \sqrt{\frac{m}{k} }$

Where:

T = period of oscillation

m = inertia mass of the oscillating body

k = spring constant

m = 2 kg , k = 2 N/m

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

$T=2\pi$

T = 6.28 s

m = 2 kg , k = 4 N/m

$T=2\pi \sqrt{\frac{2}{4} }$

$T=2\pi \sqrt{\frac{1}{2} }$

T = 4.44 s

m = 4 kg , k = 2 N/m

$T=2\pi \sqrt{\frac{4}{2} }$

$T=2\pi \sqrt{2 }$

T = 8.89 s

m = 1 kg , k = 4 N/m

$T=2\pi \sqrt{\frac{1}{4} }$

$T=\pi$

T = 3.14 s

Therefore the rank the periods of oscillation for the mass–spring systems from largest to smallest is:

m = 4 kg , k = 2 N/m (T = 8.89 s)
m = 2 kg , k = 2 N/m (T = 6.28 s)
m = 2 kg , k = 4 N/m (T = 4.44 s)
m = 1 kg , k = 4 N/m (T = 3.14 s)

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4 0

2 years ago

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What electric force would a stationary 3.8 C charge experience if it were far away from any other charges

MAVERICK [17]

Answer:

The electric force will be 0 N

Explanation:

From the question we are told that

The magnitude of the charge is $q_1 = 3.8 \ C$

Generally from Coulombs law the electric force between two charges is mathematically represented as

$F = \frac{ k * q_1 q_2 }{r^2}$

Here r is the distance of separation between that two charges.

Now from the question we are told that the charge is far away from any other charge hence we can say that the distance between the charge and any other charge is $r = \infty$

So

$F = \frac{ k * 3.8 * q_2 }{\infty^2}$

=> $F =0 \ N$

Hence the electric force will be 0 N

3 0

2 years ago

A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do

Flauer [41]

<h3>It takes 60 seconds to do the work</h3>

<em><u>Solution:</u></em>

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

$work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule$

<em><u>Find the time taken</u></em>

$power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second$

Thus it takes 60 seconds to do the work

3 0

2 years ago

A recipe for a sweet tea calls for 3 cups of sugar when you place a sugar into the tea it doesn't dissolve. How could you use di

barxatty [35]

Answer: You could dissolve it by heating it back up, then just cooling it down again.

Hope that helps!

6 0

2 years ago

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Is altitude abiotic or biotic or neither SCIENCE WORK NEEDED BY 11:59 HELP HELP HLEP

trapecia [35]

Answer:

abiotic

Explanation:

goggle:)))))))

6 0

2 years ago