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Hitman42 [59]
2 years ago
10

An object attached to an ideal spring executes simple harmonic motion. If you want to double its total energy, you could An obje

ct attached to an ideal spring executes simple harmonic motion. If you want to double its total energy, you could double the mass. double the force constant (spring constant) of the spring. double both the mass and amplitude of vibration. double the amplitude of vibration. double both the amplitude and force constant (spring con
Physics
1 answer:
olchik [2.2K]2 years ago
4 0

Answer:

double the force constant

Explanation:

For this exercise we see that the unit mass to the spring executes a simple harmonic motion, the maximum mechanical energy is

             Em = ½ k A²

where A is the range of motion

To double the energy we must increase the amplitude or the spring constant

Therefore, to double the total energy we must increase the amplitude by an amount √2

or we can increase the spring constant to double

The correct answer is to double the force constant

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3 years ago
.A cart rolling down an incline for 5.0 seconds has an acceleration of 1.6 m/s2. If the cart has a beginning speed of 2.0 m/s, a
Ilia_Sergeevich [38]

Use the formula,

\Delta x=v_it+\dfrac12at^2

where \Delta x is the cart's displacement (from the origin), v_i is its initial speed, a is its acceleration, and t is time.

\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2

\implies\boxed{\Delta x=30\,\mathrm m}

Alternatively, since acceleration is constant, we have

\dfrac{v_f+v_i}2=\dfrac{\Delta x}t

That is, we have these two equivalent expressions for average velocity, where v_f is the cart's final velocity. Solve for \Delta x:

\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}

\implies\boxed{\Delta x=30\,\mathrm m}

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3 years ago
Air entering the potential space of the pleural cavity is called
PolarNik [594]
Air  entering the  potential space of  the  pleural cavity is  called  pneumothorax. 
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3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. You reaction time
bearhunter [10]

Answer:

Explanation:

Discount the time here; it's not important. It doesn't tell you how long it takes the car to stop, it only refers to reaction time, which means nothing in the scheme of things.

The useful info is as follows:

initial velocity = 20 m/s

final velocity = 0 m/s

a = -10 m/s/s

and we are looking for the displacement. Use the following equation:

v^2=v_0^2+2aΔx

where v is the final velocity, v₀ is the initial velocity, a is the deceleration (since it's negative), and Δx is displacement. Filling in:

0^2=(20)^2+2(-10)Δx and

0 = 400 - 20Δx and

-400 = -20Δx so

Δ = 20 meters

3 0
3 years ago
A charge q is at the point x = 5.0 m , y = 0. Write expressions for the unit vectors you would use in Coulomb's law if you were
Flauer [41]

Answer:

A) \^r = \^y

B) \^r = -\^x

C) \^r = -0.14\^x + 0.99\^y

Explanation:

Coulomb's Law is

\vec{F}_{ab} = K\frac{q_1q_2}{r^2}\^r

where r is the distance between the two point charges.

The question clearly asks the unit vector expressions of the force that q exerts on the other charge. So, we do not need to find the force, but only the distance vector, r, between charges, and then we can derive the unit vector pointing only the direction of the force.

A) \vec{r} = \vec{r}_b - \vec{r}_a = (5\^x + \^y) - (5\^x + 0) = \^y\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{\^y}{1} = \^y

B) \vec{r} = \vec{r}_b - \vec{r}_a = (0 + 0) - (5\^x + 0) = -5\^x\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{-5\^x}{5} = -\^x

C) \vec{r} = \vec{r}_b - \vec{r}_a = (4.5\^x + 3.5\^y) - (5\^x + 0) = -0.5\^x + 3.5\^y\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{-0.5\^x + 3.5\^y}{\sqrt{(0.5)^2+(3.5)^2}} = \frac{-0.5\^x + 3.5\^y}{3.53} = -0.14\^x + 0.99\^y

7 0
2 years ago
Read 2 more answers
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