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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.
Physics
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

7 0
3 years ago
What is the station's orbital speed? the radius of earth is 6.37×106m, its mass is 5.98×1024kg.
Paha777 [63]

Answer:

v_o=2503.08\frac{m}{s}

Explanation:

Orbital velocity is the speed that a body that orbits around another body must have, for its orbit to be stable. For orbits with small eccentricity and when one of the masses is almost negligible compared to the other mass, like in this case, the orbital speed is given by:

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}

3 0
3 years ago
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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

v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s

7 0
3 years ago
Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2
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

Therefore 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

8 0
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Who yall think gonna win Seahawks or Vikings? Seahawks all the way
myrzilka [38]

Answer:

Seahawks

Explanation:

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3 0
3 years ago
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