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tino4ka555 [31]
3 years ago
5

How much heat is required to raise the temperature of 100 g of water from 5 degree celsius to 90 degree celsius?​

Physics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Explanation:

1-How many moles of NazCOs are in 10.0 ml of a 2.0 M solution?

2-How many moles of NaCl are contained in 100.0 ml of a 0.20 M solution?

3- What weight (in grams) of H2SO4 would be needed to make 750.0 ml of

2.00 M solution?

4-What volume (in ml) of 18.0 M H2SO4 is needed to contain 2.45 g H2S04?

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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.

 Vertical motion:

          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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Sometime around 2024, astronomers at the European Southern Observatory hope to begin using the E-ELT (European Extremely Large T
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Expression for angular resolution is;

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See attached images

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