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Nina [5.8K]
3 years ago
9

A 190 g rock at 20°C is completely immersed in 600 g of water at 80°C. Which one of the following statements is true right after

the rock goes in?
A. The water gives the rock thermal energy and gets no thermal energy in return.
B. The rock and the water give each other thermal energy, but the water gives more than it gets.
C. The rock gives the water thermal energy and gets no thermal energy in return.
D. It depends on the specific heat of the rock.
E. The rock and the water give each other thermal energy, but the rock gives more than it gets
Physics
1 answer:
Umnica [9.8K]3 years ago
8 0

Rock is completely immersed in hot water. By the second law of thermodynamics, thermal energy or heat is transferred from substance with higher temperature to substance with lower temperature until they come to thermal equilibrium i.e. both at same temperature.

It is given here that rock is at 20°C which is at lower temperature than water at 80°C. ∴Heat or thermal energy flows from water to rock. So, right choice is-

A. The water gives the rock thermal energy and gets no thermal energy in return.

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A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go
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<u>Answer:</u>

Golf ball will go a maximum of 270.36 meter.

<u>Explanation:</u>

  Projectile motion has two types of motion Horizontal and Vertical motion.

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          We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

          Considering upward vertical motion of projectile.

          In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

         0 = u sin θ - gt

          t = u sin θ/g

     Total time for vertical motion is two times time taken for upward vertical motion of projectile.

     So total travel time of projectile = 2u sin θ/g

Horizontal motion:

   We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

   In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

  So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

  Now in the given problem

     A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

     u = 51.5 m/s, for maximum range θ = 45⁰

   So maximum distance reached = \frac{51.5^2sin(2*45)}{9.81}=270.36 meter

So it will go a maximum of 270.36 meter.

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