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3 years ago

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An object is hung on the end of a vertical spring and is released from rest with the spring 3 unstressed. If the object falls 3.

42 cm before first coming to rest, find the period of the resulting oscillatory motion

1 answer:

den301095 [7]3 years ago

8 0

Answer:

The period of the resulting oscillatory motion is 0.20 s.

Explanation:

Given that,

Spring constant $k= 3\ N/m^2$

We need to calculate the time period

The object is at rest and has no elastic potential but it does has gravitational potential.

If the object falls then the the gravitational potential change in to the elastic potential.

So,

$mgh=\dfrac{1}{2}kh^2$

$m=\dfrac{1}{2}\times\dfrac{kh}{g}$

Where,h = distance

k = spring constant

Put the value into the formula

$m=\dfrac{1\times3\times3.42\times10^{-2}}{2\times9.8}$

$m=5.235\times10^{-3}\ kg$

Using formula of time period

$T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{5.235\times10^{-2}}{3}}$

$T=0.20\ sec$

Hence, The period of the resulting oscillatory motion is 0.20 s.

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Please answer the question in the picture, worth 25 points

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Answer:

Option C, increases and decreases

Explanation:

When an object making noise approaches you, the wave frequency increases leading to a higher pitch. Conversely, when it moves away from you or retreats, the wave frequency decreases leading to a lower pitch. This can be observed in ambulance sirens.

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2 years ago

A soccer ball is kicked and left

Vedmedyk [2.9K]

Answer:

Explanation:

Considering that this is parabolic motion, we know that the time the ball is in the air begins the instant it leaves the ground, reaches up to its max height, and then begins falling until it reaches the ground. Duh, right? Some important things happen during this trip. There are a few things we need to know in order to even begin the problem. Parabolic motion has x and y coordinates because it is 2-dimmensional; the acceleration in the x dimension is not the same as the acceleration in the y dimension; the velocity of an object at its max height is always 0; the time it takes to reach its max height (where the max height is half the distance the object travels) is half the time it takes to make the whole trip. Yikes. That's a lot to know and much to remember! Don't you just LOVE physics!?

For a. the hang time is the time the ball was in the air. Some of that stuff we talked about above is pertinent to solving this problem. We know that the velocity of the ball is 0 at its max height, and we also know that if we find the time it takes to reach its max height, we can double that number to find how long it was in the air for the whole trip. Use the one-dimensional equation

$v=v_0+at$ to find out how long it took to reach the max height. Even though we don't yet know the max height, we DO know that the velocity at that point is 0. BUT before we do that, since we are working in the y-dimension only, it would behoove us (benefit us) to find the velocity particular to this dimension. We are going to answer c. first, then backtrack.

c. wants the initial vertical velocity. That is found in the magnitude of the "blanket" or generic velocity times the sin of the angle, namely:

$V_y=25sin(45)$ so

$V_y=$ 18 m/s Now we can use that as the initial upwards velocity in part a:

$v=v_0+at$ and filling in:

0 = 18 + (-9.8)t and

-18 = -9.8t so

t = 1.8 seconds. But remember, this is only half the time it was in the air. The whole trip, then, takes 2(1.8) which is

t = 3.6 seconds

That's a and c. Now for b:

b. asks for the x component of the velocity:

$V_x=Vcos\theta$ which works out to be the same as the vertical velocity, since the sin and cos of 45 degrees is the same:

$V_x=25cos45$ and

$V_x=$ 18 m/s

Onto d:

d. wants the max height. Remember, it took 1.8 seconds to get to the max height, so using yet another one-dimensional equation:

Δx = v₀t + $\frac{1}{2}at^2$ where Δx is the displacement, v₀ is the initial upwards velocity, a is the pull of gravity, and t is the time it takes to reach that max height (Δx, our unknown). Filling in:

Δx = $18(1.8)+\frac{1}{2}(-9.8)(1.8)^2$ and if you do the rounding correctly, you'll end up with this:

Δx = 32 - 16 so

the max height, Δx, is 16 meters.

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Δx = vt so

Δx = 18(3.6) remember that the ball was in the air for a total of 3.6 seconds, so

Δx = 65 meters.

Phew!!!!! That's a lot! I suggest you learn your physics or this will make you insane by the end of the course!

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3 years ago

If you hold a bar magnet in each hand and bring your hands together, will the force be attractive or repulsive if the magnets ar

Law Incorporation [45]

a) The force is repulsive

b) The force is attractive

Explanation:

Every magnet has a magnetic field around it. It is possible to distinguish two different poles in the magnet, according to the direction of the magnetic field: in particular, the lines of the field go out from the North pole and go into the South Pole. When a magnet is broken, two new magnets are formed, each of them having its own north and south pole.

The force between two magnets can be either attractive or repulsive, depending on which poles are facing each other. We have the following situation:

The magnetic force between two like poles (north-north and south-south) is repulsive
The magnetic force between two opposite poles (north-south) is attractive

Therefore, we have the following situations in this problem:

a)

Here we are holding the two north poles together: since they are like poles, they repel each other, so the force in this case is repulsive

b)

Here we are holding a north pole and a south pole together: since they are opposite poles, they attract each other, so the force in this case is attractive

Learn more about magnetic fields:

brainly.com/question/3874443

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3 years ago

What distance separates Earth and the Sun? A 300,000 km per second B Eight light-minutes C 149 million light-seconds D One light

lina2011 [118]

Answer:

B Eight light-minutes

Explanation:

In the case when the distance separated earth and the sun so here we orbit the sun for a 150 million km distance and the light moves would be 300,000 kilometers per second

Now divide this

= 150 million ÷ 300,000 kilometers per second

= 500 seconds

This 500 seconds represent 8 minutes and 20 seconds

Hence, option B is correct

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2 years ago

Read 2 more answers

Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same

Contact [7]

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

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$\underset{V_{B}}{\rightarrow}$ = velocity of car B before collision = $V_{B}$ i + 0 j

$\underset{V_{AB}}{\rightarrow}$ = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

$M_{A}$ = mass of car A = 1750 kg

$M_{B}$ = mass of car B = 1450 kg

Using conservation of momentum

$M_{A}$ $\underset{V_{A}}{\rightarrow}$ + $M_{B}$ $\underset{V_{B}}{\rightarrow}$ = ( $M_{A}$ + $M_{B}$ ) ( $\underset{V_{AB}}{\rightarrow}$ )

(1750) (0 i - $V_{A}$ j) + (1450) ( $V_{B}$ i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) $V_{B}$ i - (1750) $V_{A}$ j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) $V_{B}$ = 97600 and - (1750) $V_{A}$ = - 60160

$V_{B}$ = 67.3 km/h and $V_{A}$ = 34.4 km/

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3 years ago