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aksik [14]
3 years ago
11

Your pet snake would be found in which kingdom:

Physics
1 answer:
Tasya [4]3 years ago
5 0
The last one is the answer hope this help on what  u r doing
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The amount of intensity and duration of sunlight striking earth vary with latitude
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( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration
Talja [164]

Answer:

4.725 m/s^{2}

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then a=\frac {F}{m}  where a is acceleration, F is force and m is mass

Also making mass the subject of the formula m=\frac {F}{a}

For m1= \frac {F}{13.5} and m2=\frac {F}{3.5} hence F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}

4 0
3 years ago
Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0
3 years ago
A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh
kolezko [41]

Answer:

The angular velocity is  w = 53.35 \ rounds /minute

Explanation:

From the question we are told that

    The mass of the block is  m = 61.2kg

     The of the pulley is  M = 14.2 kg

      The radius of the pulley is  R = 1.5m

       The radius  of the cord around the pulley is  r = 1.5 m

       The distance of the block to the floor is  d = 8.0 m

         

From the question we are told that the moment of inertia of the pulley is

          I  = \frac{1}{2} MR^2 kg \cdot m^2

Substituting value  

         I = \frac{1}{2}  * 14.2 * (1.5)^2

         I = 15.975 kg \cdot m^2

Using the Newtons law we can express the force acting on the vertical axis as

              ma = mg -T

         =>  T = mg -ma

Now when the pulley is rotated that  torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

                  \tau = I \alpha

     Here \alpha is the angular acceleration

           Here \tau is the torque which can be equivalent to

              \tau = T r

  Substituting this above

            Tr = I \alpha      

Substituting for T

         (mg - ma ) r =  I\  r \alpha

Here a is the  linear acceleration which is mathematically represented as

           a = r\alpha

    (mg - m(r\alpha ) ) r =  I\  r \alpha

     mgr = I\alpha  + m(r\alpha ) r

    mgr = \alpha  [ I + mr^2]

   making \alpha the subject

          \alpha  = \frac{mgr}{I -mr ^2}          

   Substituting values

            \alpha  = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}

             \alpha =5.854 rad /s^2

Now substituting into the equation above to obtain the acceleration

             a = 5.854 * 1.5

                a=8.78 m/s^2

This acceleration is a = \frac{v}{t}

and v is the linear velocity with is mathematically represented as

         v = \frac{d}{t}

Substituting this into the formula acceleration

        a = \frac{d}{t^2}

making t the subject

         t = \sqrt{\frac{d}{a} }

substituting value

      t = \sqrt{\frac{8}{8.78}}

     t = 0.9545 \ s

Now the linear velocity is

       v = \frac{8}{0.9545}

       v = 8.38 m/s

The angular velocity is  

       w = \frac{v}{r}

So

       w = \frac{8.38}{1.5}

        w = 5.59 rad/s

Generally 1 radian is equal to  0.159155 rounds or turns

        So  5.59 radian is  equal to x

Now x is mathematically obtained as

         x = \frac{5.59 * 0.159155}{1}

            = 0.8892 \ rounds

 Also

      60  second =  1 minute

So   1 second  = z      

Now z is mathematically obtained as

         z = \frac{ 1}{60}

            z = 0.01667 \ minute

Therefore

              w = \frac{0.8892}{0.01667}

              w = 53.35 \ rounds /minute

           

8 0
3 years ago
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