What we crystalline metals?

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

4 years ago

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Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Reika [66]

Answer:

v = 21 m / s

Explanation:

We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data

v = v₀ + a t

the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s

a = (v-v₀) / t

a = (11 - 6) / 8

a = 0.625 m / s²

when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s

v = v₀ + a t

v = 11 + 0.625 16

v = 21 m / s

4 0

4 years ago

How much electrical potential energy is stored in a capacitor that has 8.0 × 10-10 C of charge on each plate and a potential dif

arsen [322]

THere is a standard relationship that gives this result where the capacity of the capacitor is used:
$E=1/2* \frac{Q^2}{C}$ .
We know though that Q/c=V and thus we can use the relationship:
E=Q*V/2 where we have just substituted in. If we also take into account that Q=VC, we can also get that E=V^2*C/2.
We are given the charge and the potential, so the best expression to use is the middle one.
Substituting, we get that E=1/2*8*10^(-10)*20=8*10^(-9).

The answer is B

8 0

3 years ago

A boat crosses a river of width 229 m in which the current has a uniform speed of 0.756 m/s. The pilot maintains a bearing (i.e.

Stolb23 [73]

Answer:

1.85 m/s

102.438 m

Explanation:

From vector motion

The resultant velocity

$R=\sqrt{0.756^2+1.69^2}\\\Rightarrow R=1.85\ m/s$

The magnitude of speed of the speed of the boat relative to a stationary shore observer is 1.85 m/s

The distance covered is 229 m

Time = Distance / Speed

$\text{Time}=\frac{229}{1.69}\\\Rightarrow \text{Time}=135.50\ s$

Distance = Speed of current × Time

Distance = 0.756×135.5 = 102.438 m

The boat will be 102.438 m from the initial position is the boat when it reaches the opposite shore

3 0

4 years ago

Use the circuit diagram to decide if the lightbulb will light. Justify your answer

elena-14-01-66 [18.8K]

Answer:

It will.

Explanation:

For a circuit to function properly, everything has to be connected. As the diagram shows that everything is connected and there are no gaps, it will light.

3 0

3 years ago

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