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nlexa [21]
3 years ago
6

Which type of energy is stored in a stretched string on the bow in the following figure ​

Physics
2 answers:
emmasim [6.3K]3 years ago
8 0

Answer:

potential energy

Explanation:

Anestetic [448]3 years ago
4 0

Answer: Elastic Potential Energy

Explanation: For stretched material such as ropes or strings they have a certain form of energy that allows them to stretch or compress. The energy is called elastic energy. For the case of the string on the bow it possess Elastic Potential Energy since there is no motion in the string.

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You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u
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first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

d_1 = v*t_1

d_1 = 10*20 = 200 m

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

d_2 = v*t = 20*20 = 400 m

so the net displacement is given as

\vec d = \vec d_2 - \vec d_1

\vec d = 400 - 200 = 200 m

so it will be displaced by total displacement 200 m

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(m/s)<br> 20<br> 10.<br> 100<br> -10<br> total distance
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sum of them 4

Explanation:

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If the refractive index of water is 1.33, then its critical angle is...
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An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec
lutik1710 [3]

Answer:

The temperature of the metal is  T_m  =  376.8 ^o C

Explanation:

From the question we are told that

     The mass of the metal is  M =  60 \ kg

     The specific heat of the metal is  c_p  =  0.1027 kcal/(kg \cdot ^oC)

       The mass of the oil is M_o  =  810 \ kg

       The temperature of the oil is  T_o  =  35^oC

       The specific heat of oil is  c_o  =  0.7167 kcal/(kg \cdot ^oC )

       The equilibrium temperature is T_e  =  39 ^oC

According to the law of energy conservation

     Heat lost by metal  =  heat gained by the oil

So  

   The quantity  of heat lost by the metal is mathematically represented as

               Q =  - Mc_p \Delta T

=>            Q =  -Mc_p (T_m  -  T_c)

Where T_ m  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

           Q =  M_o c_o \Delta T

=>        Q =  M_o c_o (T_c - T_o)

So  

         Mc_p (T_m  -  T_c)   =   M_o c_o (T_c - T_o)

substituting values

          - 60 * 0.1027 (T_m  - 39)   =   810 * 0.7167 *  (39 - 35)

=>       T_m  =  376.8 ^o C

         

6 0
3 years ago
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