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Igoryamba
2 years ago
8

A mass of (200 of hot water at (75, 0 degrees * C) is mixed with cold water of mass M at (5, 0 degrees * C) The final temperatur

e of the mixture is (25, 0 degrees * C) What is the mass of the cold water (M)
Physics
2 answers:
RUDIKE [14]2 years ago
6 0

∆T for hot water

  • 75-25
  • 50°C

∆T for cold water

  • 25-5
  • 20°C

Both are equal according law of conservation of energy

\\ \rm\Rrightarrow m_1c\delta T_0=m_2c\delta

\\ \rm\Rrightarrow 200(50)=m_2(20)

\\ \rm\Rrightarrow 10000=m_2(20)

\\ \rm\Rrightarrow m_2=500g=0.5kg

Ratling [72]2 years ago
3 0

Q in = Q out

mcΔt in = mcΔt out (c = equal)

mΔt in = mΔt out

200 x  (75-25) = M x (25-5)

M = 500 g

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0 m/s

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3 years ago
A 12 kg<br> mass is lifted to a height of 2 m. What is its potential energy<br> at this position?
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Answer:

Explanation:

Potential energy is the energy stored within an object, due to the object's position, arrangement or state

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2 years ago
Read 2 more answers
A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. Yo
aniked [119]

Answer:

a) 46.5º  b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

     θ = 43.5º

This angle measured from the surface is

     θ_r = 90 -43.5

     θ_s = 46.5º

b) In this part the law of refraction must be used

     n₁ sin θ₁ = n₂. Sin θ₂

     sin θ₂ = n₁ / n₂ sin θ₁

The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

     θ_s = 90 - θ

     θ_s = 90- 43.5

     θ_s = 46.5º

     sin θ₂ = 1 / 1.68  sin 46.5

     sin θ₂ = 0.4318

     θ₂ = 25.6º

The angle with respect to the surface is

     θ₂_s = 90 - 25.6

     θ₂_s = 64.4º

measured in the fourth quadrant

3 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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Akimi4 [234]
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