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Manufactured bolts can’t be too long or too short or they might not fit where they need to. Help determine which bolts will work

Sever21 [200]

What is the longest the bolt can be and still be acceptable

5 0

3 years ago

While spinning down from 500 rpm to rest, a flywheel does 3.9 kj of work. this flywheel is in the shape of a solid uniform disk

Ksivusya [100]

The answer is 4.0 kg since the flywheel comes to rest the kinetic energy of the wheel in motion is spent doing the work. Using the formula KE = (1/2) I w².

Given the following:

I = the moment of inertia about the axis passing through the center of the wheel; w = angular velocity ; for the solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.

The radius = 1.2 m and KE = 3900 J

3900 J = (1/4) m (1.2)² (52.36)²

m = 3900 J / (0.25) (1.2)² (52.36)²

m = 3.95151 ≈ 4.00 kg

6 0

4 years ago

Science-<br> is there such thing as an altocirrus cloud?<br> yes or no?

atroni [7]

Ofcourse the answer is yes

7 0

3 years ago

Read 2 more answers

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

marusya05 [52]

(a) $7.32\cdot 10^7 Hz$

The frequency of an electromagnetic waves is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=4.1 m$ is the wavelength of the wave in the problem

Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

(b) $4.60\cdot 10^8 rad/s$

The angular frequency of a wave is given by

$\omega = 2\pi f$

where

f is the frequency

For this wave,

$f=7.32\cdot 10^7 Hz$

So the angular frequency is

$\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s$

(c) $1.53 m^{-1}$

The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

where

$\lambda$ is the wavelength of the wave

For this wave, we have

$\lambda=4.1 m$

so the angular wave number is

$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

(d) $1.03\cdot 10^{-6}T$

For an electromagnetic wave,

$E=cB$

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$