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Dennis_Churaev [7]
3 years ago
15

Explain the correspondence that lets us easily translate between linear motion and rotational motion. What are the linear analog

ues of the rotational quantities we have discussed in lecture (i.e. angle, angular velocity, angular acceleration, moment of inertia)? Where does the correspondence seem to fail?
Physics
1 answer:
RideAnS [48]3 years ago
4 0

Explanation:

The linear analog of angle is angle itself.

The linear analog of angular velocity is linear velocity.

ω is angular velocity, therefore linear velocity is given by v

∴ for linear velocity, v^{2} = u^{2}+2.a.S

   for angular velocity, \omega_{f}^{2}  = \omega _{i}^{2}+2.a.S

The linear analog of angular acceleration is acceleration.

α is angular acceleration whereas as a is linear acceleration.

∴ for linear acceleration, v = u + a.t

  for angular acceleration, \omega_{f}= \omega _{i}+\alpha .t

The linear analog of moment of inertia is mass.

I is moment of inertia and m is mass,

∴ for linear analog, F = m.a

  for angular analog, τ - I.α

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Explanation:

5 0
3 years ago
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.
kodGreya [7K]

Answer:

The  z-component of the force is  \= F_z  =  0.00141 \ N    

Explanation:

From the question we are told that

          The charge on the particle is q =  7.76 *0^{-8} \  C    

           The magnitude of the magnetic field is  B =  0.700\r i \ T

            The  velocity of the particle toward the x-direction is  v_x  =  -1.68*10^{4}\r  i  \ m/s

           The  velocity of the particle toward the y-direction is

v_y  =  -2.61*10^{4}\ \r j  \ m/s

           The  velocity of the particle toward the z-direction is

v_y  =  -5.85*10^{4}\ \r k  \ m/s

Generally the force on this particle is mathematically represented as

          \= F  =  q (\= v   X  \= B )

So  we have    

          \= F  =  q ( v_x \r  i + v_y \r  j  +  v_z \r k  )  \ \ X \ (  \= B i)

         \= F  = q (v_y B(-\r  k) + v_z B\r j)      

  substituting values

       \= F  = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r  z) + [(5.58*10^{4}) (0.700)]\r y)    

      \= F=  0.00303\ \r j +0.00141\ \r k                  

So the z-component of the force is  \= F_z  =  0.00141 \ N    

Note :  The  cross-multiplication template of unit vectors is  shown on the first uploaded image  ( From Wikibooks ).

7 0
3 years ago
Mass is a measure of weight. True False
8_murik_8 [283]

Answer: False

Explanation: Mass is a measure of the amount of matter in an object.

5 0
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Sedimentary rock I believe
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