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Hatshy [7]
2 years ago
6

A ceiling fan has five blades, each with a mass of 0.34 kg and a length of 0.66 m. The fan is operating in its "low" setting at

which the angular speed is 9.4 rad/s. If the blades can be approximated as uniform thin rods that rotate about one end, what is the total rotational kinetic energy of the five blades?

Physics
1 answer:
expeople1 [14]2 years ago
8 0

Answer: The total rotational kinetic energy of the five blades is 10.9J

Explanation: Please see the attachments below

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In a distant solar system, a giant planet has
sergeinik [125]

Answer:

mass of the planet: 5.9\,10^{26}\,kg

Explanation:

When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

m\frac{v^2}{R}=G\frac{M\,m}{R^2}

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R (5.32\,10^5\,km=5.32\,10^8\,m, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference (2\,\pi\,R) in 388800 seconds, therefore its tangential velocity is:

v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg

8 0
2 years ago
A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit
Maslowich

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= -  GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = \frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}

=  - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}

=   - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

8 0
2 years ago
A frog jumps to the left with an average speed of
Bingel [31]

Answer:

<h3>0.99 m</h3>

Explanation:

Average velocity is the change of rate of displacement with respect to time;

Average velocity = Displacement/Time

Given

Average velocity of the frog = 1.8m/s

Time = 0.55s

Required

Displacement of the frog

Substitute the given parameters into the formula;

1.8 = displacement/0.55

cross multiply

Displacement = 1.8*0.55

Displacement = 0.99 m

Hence the frog's displacement is 0.99m

7 0
1 year ago
(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang
avanturin [10]

6m at 0.9 radian means 6m at 51.57^0

Since positive angles are counter clockwise

6m at 51.57^0 can be written as 6*cos51.57^0i+6*sin51.57^0j = 3.73i+4.70j

5m at -75^0 can be written as 5*cos(-75)^0i+5*sin(-75)^0j = 1.294i-4.83j

4m at 1.2 radian means 4m at 68.75^0

4m at 68.75^0 can be written as 4*cos68.75^0i+4*sin68.75^0j = 1.45i+3.73j

6m at -210^0 can be written as 6*cos(-210)^0i+6*sin(-210)^0j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = \sqrt{1.274^2+0.6^2} = 1.98 m

c) Angle,  tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0

7 0
1 year ago
An organ pipe is tuned to exactly 384 Hz when the temperature in the room is 20°C. Later, when the air has warmed up to 25°C, th
maksim [4K]

Answer: A. Greater than 384 Hz

Explanation:

The velocity of sound is directly related to the temperature rather it is directly proportional meaning if the temperature decreases the velocity decreases and if temperature increases the velocity increases.

Now, we are given that temperature has risen from 20°C to 25°C meaning it has increases. So it implies that velocity must also increase.

Also, the velocity for organ pipe is directly proportional to its frequency. Now if velocity increases frequency must also increase. In this case, the original frequency is 384 Hz. Now increasing the temperature resulted in increase in velocity and thus increase in frequency.

So option a is correct. i.e. now frequency will be greater than 384 Hz.

3 0
2 years ago
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