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crimeas [40]
3 years ago
12

A 0.18 m radius pulley is free to rotate about a horizontal axis. A mass and a mass are attached by a massless string, which is

hung over the pulley. If the string does not slip, calculate the magnitude of the net torque on the pulley about its rotational axis.
Physics
1 answer:
Musya8 [376]3 years ago
8 0

Answer:

T = 1.766(M-m) Nm where M and m are the 2 masses of the objects

Explanation:

Let m and M be the masses of the 2 objects and M > m so the system would produce torque and rotational motion on the pulley. Force of gravity that exert on each of the mass are mg and Mg. Since Mg > mg, the net force on the system is Mg - mg or g(M - m) toward the heavier mass.

Ignore friction and string mass, and let g = 9.81 m/s2, the net torque on the pulley is the product of net force and arm distance to the pivot point, which is pulley radius r = 0.18 m

T = Fr = g(M - m)0.18 = 0.18*9.81(M - m) = 1.766(M-m) Nm

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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
An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.
Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis  E=31.25\ N/c and E = 15.625\ N/c respectively.

Explanation:

Given :

Distance between charges , d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.

Magnitude of charges , q=1\ nC = 10^{-9}\ C.

Dipole moment , p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

E=\dfrac{2kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E=31.25\ N/c.

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

E = \dfrac{kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E = 15.625\ N/c.

Hence , this is the required solution.

7 0
3 years ago
How is displacement different from distance?
Anni [7]

Answer:

i do belive its C

Explanation:

i remeber this question from somewhere also it makes the most sense

5 0
3 years ago
Read 2 more answers
Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.
pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is  D = 0.59 \  m    

Explanation:

From the question we are told that

      The best resolution is  \theta  =  0.3 \  arcsecond

       The  wavelength is  \lambda  =  700 \  nm =  700 *10^{-9 } \  m

       

Generally the

         1 arcminute  = >  60 arcseconds

=>      x arcminute =>   0.3 arcsecond

So

       x =  \frac{0.3}{60 }

=>    x = 0.005 \  arcminutes

Now

         60 arcminutes  =>  1 degree

          0.005 arcminutes = >  z degrees  

=>       z =  \frac{0.005}{60 }

=>      z =  8.333 *10^{-5}  \ degree

Converting to radian  

           \theta  = z =  8.333 *10^{-5}  * 0.01745 = 1.454 *10^{-6} \  radian

Generally the resolution is mathematically represented as

            \theta  =  \frac{1.22 *  \lambda  }{ D}

=>    D =  \frac{1.22 * \lambda }{\theta }

=>     D =  \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }    

=>     D = 0.59 \  m    

4 0
3 years ago
cup of hot chocolate has a spoon in it. How does the heat move from the hot chocolate to the metal spoon handle?
serg [7]

The heat moves from the hot chocolate to the handle of the spoon by a process called thermal conduction.

It is the transfer of heat energy from one object to another when they are in contact with eachother.

Hope this answers your question.

5 0
3 years ago
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