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Lubov Fominskaja [6]

3 years ago

12

A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the ki

netic energy of the fan after 8.0 s.

1 answer:

Mars2501 [29]3 years ago

4 0

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

$\tau=I*\alpha$

given

$\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?$

$\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2$

the angular velocity is

$\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s$

$KE= 1/2*I* \omega^2$

$KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J$

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What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together?

myrzilka [38]

<h2>Answer:</h2>

786Ω and 20.32Ω respectively.

<h2>Explanation:</h2>

(a) Given a number of resistors each with its own resistance, the largest resistance can be obtained when these resistors are connected in series.

From the question, the resistors have the following resistances;

36.0-Ω, 50.0-Ω , and 700-Ω

Now, when they are connected in series, the total resistance (R) obtainable is given by the sum of these individual resistances as follows;

R = 36.0-Ω + 50.0-Ω + 700-Ω

R = 786Ω

Therefore, the largest resistance that can be obtained by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 786Ω

(b) Similarly, given a number of resistors each with its own resistance, the smallest resistance can be obtained when these resistors are connected in parallel.

From the question, the resistors have the following resistances;

36.0-Ω, 50.0-Ω , and 700-Ω

Now, when they are connected in parallel, the total resistance (R) obtainable is given by using the relation as follows;

$\frac{1}{R}$ = $\frac{1}{36.0}$ + $\frac{1}{50.0}$ + $\frac{1}{700.0}$

$\frac{1}{R}$ = $\frac{35000+25200+1800}{1260000}$

$\frac{1}{R}$ = $\frac{62000}{1260000}$

$\frac{1}{R}$ = $\frac{62}{1260}$

R = $\frac{1260}{62}$

R = 20.32Ω

Therefore, the smallest resistance that can be obtained by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 20.32Ω

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Answer:

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Explanation:

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$v_o=\sqrt{\frac{GM}{r}}$

Where M is the greater mass around which this negligible body is orbiting, r is the radius of the greater mass and G is the universal gravitational constant. So:

$v_o=\sqrt{\frac{6.674*10^{-11}\frac{m^3}{kg\cdot s^2}(5.98*10^{24}kg)}{6.37*10^6m}}\\v_o=2503.08\frac{m}{s}$

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(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

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Alona [7]

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

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