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3 years ago

15

An old clock has a spring that must be wound to make the clock hands move. Which statement describes the energy of the spring an

d hands of the clock?
A)
The spring has kinetic energy and the hands have kinetic energy.

B)
The spring has kinetic energy and the hands have potential energy.

C)
The spring has potential energy and the hands have potential energy.

D)
The spring has potential energy and the hands have kinetic energy.

1 answer:

topjm [15]3 years ago

8 0

Answer:

c

Explanation:

neither the spring or hands are in the action of movemnet

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scoundrel [369]

Answer:

<u>Amplitude - remains the same</u>

<u>Frequency - increases</u>

<u>Period - decreases</u>

<u>Velocity - remains the same.</u>

<u />

Explanation:

The amplitude of the wave remains the same since you are not changing the distance your hand moves and the amplitude of the wave depends on how much distance your hand covers while moving.

The frequency of your wave increases since now you are moving your hand more number of times in the same period i.e. your hand is moving faster in one second. So, the frequency of your wave increases.

The period is the time taken by the wave to travel a certain distance. Since your hand is now moving faster, the wave will travel faster and will take less time to cover the same distance hence, we can say that its period will decrease.

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2 years ago

Please help me with this question

valkas [14]

Answer: m∠P ≈ 46,42°

because using the law of sines in ΔPQR

=> sin 75°/ 4 = sin P/3

so ur friend is wrong due to confusion between edges

+) we have: sin 75°/4 = sin P/3

=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16

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Explanation:

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3 years ago

If displacement covered by a particle is zero then distance cover by it<br>

sleet_krkn [62]

Answer:

When displacement is zero, the particle may be at rest, therefore, distance travelled = 0.

Again, when displacement is zero, the final position matches with the initial position after some time, but the distance travelled will not be zero.

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2 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

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2 years ago

Which is not a common property of ionic compounds?

Dmitriy789 [7]

Answer:

low melting point

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Explanation:

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2 years ago