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Phantasy [73]
3 years ago
10

When a solid uniform sphere is spinning about an axis of rotation through its center, its rotational kinetic energy is K and mom

ent of inertia I = ⅖ MR2. A second solid sphere having twice the mass and twice the diameter of the first one is spinning about an axis through its center and has a twice the angular velocity of the first sphere. The rotational Kinetic energy of the second sphere is:
8K
4K
32K
2K
K
Physics
2 answers:
expeople1 [14]3 years ago
7 0
32 kg is rightttttt oneeeee!!!!!
lawyer [7]3 years ago
5 0

Answer:

uh.

Explanation:

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A 1.55-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 13.3-n hori
yarga [219]
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.


First, the vertical component of tension (Tsin theta) is equal to the weight of the object. 
 T * sin θ = mg =</span> 1.55 * 9.81 <span>
 T * sin θ = 15.2055

Second, the horizontal component of tension (t cos theta) is equal to the force of the wind. 
 T * cos θ =  13.3 

Tan θ = sin  </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82. 

T then is equal to 20.20 N
4 0
3 years ago
You tie a cord to a pail of water and swing the pail in a vertical circle of radius 0.710 mm . What minumum speed must the pail
Blababa [14]

Answer:

The minumum speed the pail must have at its highest point if no water is to spill from it

= 2.64 m/s

Explanation:

Working with the forces acting on the water in the pail at any point.

The weight of water is always directed downwards.

The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.

And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.

At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.

Net force = W + (normal force)

But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.

At this point of minimum velocity,

Normal force = 0

Net force = W

Net force = centripetal force = (mv²/r)

W = mg

(mv²/r) = mg

r = 0.710 m

g = 9.8 m/s²

v² = gr = 9.8 × 0.71 = 6.958

v = √(6.958) = 2.64 m/s

Hope this Helps!!!

7 0
3 years ago
Which feature of the sun appears in cycles of about 11 years?
goldfiish [28.3K]
The Sun's magnetic field goes through a cycle, called the solar cycle. Every 11 years or so, the Sun's magnetic field completely flips. This means that the Sun's north and south poles switch places. Then it takes about another 11 years for the Sun's north and south poles to flip back again.
8 0
4 years ago
The answer to number 9 please
Delicious77 [7]
The magnitude is doubled. The direction doesn't change.
7 0
3 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
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