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2 years ago

Potential energy is based on an object's

2 answers:

6 0

Potential energy is when an object is in position, means when an object isn’t moving at all its like rested, therefore the answer is position, C.

VashaNatasha [74]2 years ago

4 0

Answer:

C. position

Explanation:

potential energy is based on stored energy, therefore it is C

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These graphs show measurements of the density of air as different sound waves pass a single point.

Ymorist [56]

Answer: D!

It is the option with the greatest amplitude.

5 0

3 years ago

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A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is broug

natali 33 [55]

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$ , assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$ .

Initial kinetic energy ( ${\rm KE}$ ) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$ .

When the vehicle is brought to a rest, the elastic potential energy ( $\text{EPE}$ ) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$ .

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$ .

Rearrange this equation to find an expression for $k$ , the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$ .

Substitute in the given values $m = 1508\; {\rm kg}$ , $v \approx 1.908\; {\rm m\cdot s^{-1}}$ , and $x = 6.87\; {\rm m}$ :

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$

8 0

1 year ago

Why does the frequency of a siren get higher as an ambulance using that siren gets closer

Nadya [2.5K]

The frequency of a wave becomes higher due to the object moving at a fast pace coming towards you with shorter wavelengths (depending on the speed) aka the Doppler Effect.
Hope this helps

3 0

2 years ago

Read 2 more answers

A periodic wave has a wavelength of 0.60 m and a speed of 40 m/s. What is the<br> wave frequency?

levacccp [35]

The correct answer is .05

4 0

2 years ago

A runner of mass 53.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cen

const2013 [10]

Answer:

0.336 rad/s

Explanation:

$\omega_1$ = Angular speed of the turntable = -0.2 rad/s

R = Radius of turntable = 2.9 m

I = Moment of inertia of turntable = $76\ kgm^2$

M = Mass of turn table = 53 kg

$v_1$ = Magnitude of the runner's velocity relative to the earth = 3.6 m/s

As the momentum in the system is conserved we have

$Mv_1R+I\omega_1=(I + MR^2)\omega_2\\\Rightarrow \omega_2=\dfrac{Mv_1R+I\omega_1}{I + MR^2}\\\Rightarrow \omega_2=\dfrac{53\times 3.6-76\times 0.2}{76+53\times 2.9^2}\\\Rightarrow \omega_2=0.336\ rad/s$

The angular velocity of the system if the runner comes to rest relative to the turntable which is the required answer is 0.336 rad/s

4 0

2 years ago