Having more ______________ will make a material a better conductor.

A Blu-ray disc is approximately 11 centimeters in diameter. The drive motor of a Blu-ray player is able to rotate up to 10,000 r

stepladder [879]

Answer:

the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates is 57.6 m/s

Explanation:

Given information:

diameter of the disc, d = 11 cm, r = 5.5 cm = 0.055 m

angular speed ω = 10000 rev/min = (10000 rev/min)(2π rad/rev)(1/60 min/s)

= 1000π/3 rad/s

to calculate the maximum angular speed we can use the following formula

ω = v/r

v = ωr

= (1000π/3)(0.055)

= 57.6 m/s

8 0

3 years ago

An LC circuit consists of a 3.14 mH inductor and a 5.08 µF capacitor. (a) Find its impedance at 55.7 Hz. 563.57 Correct: Your an

ANEK [815]

Answer:

a)

$z=561.7$

b)

$z=214.1$

Explanation:

L = inductance of the Inductor = 3.14 mH = 0.00314 H

C = capacitance of the capacitor = 5.08 x 10⁻⁶ F

a)

f = frequency = 55.7 Hz

Impedance is given as

$z=\frac{1}{2\pi fC} - 2\pi fL$

$z=\frac{1}{2(3.14) (55.7)(5.08\times 10^{-6})} - 2(3.14) (55.7)(0.00314)$

$z=561.7$

b)

f = frequency = 11000 Hz

Impedance is given as

$z= - \frac{1}{2\pi fC} + 2\pi fL$

$z= - \frac{1}{2(3.14) (11000)(5.08\times 10^{-6})} + 2(3.14) (11000)(0.00314)$

$z=214.1$

4 0

3 years ago

How to read a vernier calipers?

RideAnS [48]

Answer:

I'll write it below

Explanation:

1) understand the parts.

2)read the scales

3)check the scale of your smallest divisions

4)clean the object you are measuring

5)If you have, unlock the screw

6)close the jaws

I hope this satisfies you sir.

If you have any questions related to this please feel free to ask me. I hope u will follow me and make this the brainliest answer.

7 0

3 years ago

Read 2 more answers

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th

tiny-mole [99]

Answer:

I=1.48 A

Explanation:

Given that

B=3.1 x 10⁻5 T

b= 4.2 cm

l= 9.5 cm

The relationship for magnetic field and current given as

$B=\dfrac{2\mu _oI}{\pi}D$

Where

$D=\dfrac{\sqrt{l^2+b^2}}{lb}$

By putting the values

$D=\dfrac{\sqrt{l^2+b^2}}{lb}$

$D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}$

D=26.03 m⁻¹

$B=\dfrac{2\mu _oI}{\pi}D$

$3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03$

$I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}$

I=1.48 A

8 0

3 years ago

A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan

Lapatulllka [165]

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

$v^2 - v_0^2 = 2as$ where v = 23 m/s is the velocity of the car when it passes the worker, $v_0$ = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

$23^2 - 0^2 = 2*a*255$

$510a = 529$

$a = 529 / 510 = 1.04 m/s^2$

8 0

3 years ago