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son4ous [18]
3 years ago
6

Question 2 (1 point)

Physics
1 answer:
sveticcg [70]3 years ago
7 0
according to the facts it should be a numerous amount of wrong answers so ignore dis
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An archer pulls a bow string 0.5 m. If the spring constant is 16,000 N/m, what is the energy stored in the bow string?
mars1129 [50]

2000J

Explanation:

Given parameters:

Extension = 0.5m

Spring constant = 16000N/m

Unknown:

Energy stored in the bow string = ?

Solution:

The energy stored in a bow string is an elastic potential energy.

It can be calculated using the expression below;

     Elastic energy = \frac{1}{2} K e²

Where k is the spring constant

            e is the extension

Input the parameters;

  Elastic energy = \frac{1}{2} K e²

                          =\frac{1}{2} x 16000 x 0.5²

                          = 2000J

learn more:

Potential energy brainly.com/question/10770261

#learnwithBrainly

3 0
3 years ago
I will GIVE BRAINLYEST AND POINTS PLEAS HELP!!!
Elenna [48]

Answer:

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Explanation:

I figured that these answers made the most sense so ya.

3 0
3 years ago
A helicopter is traveling with a velocity of 12 m/s directly upward.
Irina-Kira [14]

Answer:

H = 27.35 m

Explanation:

From projectile motion, we know that;

h = (v_o)²/2g

Where;

v_o = 12 m/s

g = 9.8 m/s²

Thus;

h = 12²/(2 × 9.8)

h = 144/19.6

h = 7.35 m

Now, we are told that when the man is 20 m above the pillow, he lets go of

the rope.

Thus, greatest height reached by the man above the ground is;

H = 20 + h

H = = 20 + 7.35

H = 27.35 m

5 0
3 years ago
1.) A ball is thrown horizontally from the top of a building. What is the vertical velocity of the ball after 1 second?
BARSIC [14]
  1. 9.8
  2. zero
  3. 45
  4. not sure - 70
  5. Uniform accelerated motion
6 0
3 years ago
Read 2 more answers
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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