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Mars2501 [29]
2 years ago
14

Brass is an alloy made from copper and zinc a 0.59 kg brass sample at 98.0 is dropped into 2.80 kg of water at 5.0 c if the equi

librium temperature is 6.8 what is the specific heat capacity
Physics
2 answers:
MariettaO [177]2 years ago
6 0

Answer:

393.399 J/kg.°C

Explanation:

Specific heat capacity: This is the quantity of heat required to raise the temperature of a unit mass of a substance through a degree rise in temperature.

Heat lost by the brass = heat gained by water

CM(t₁-t₃) = cm(t₃-t₂)........................ Equation 1

Where C = specific heat capacity of the brass, M = mass of the brass, t₁ = initial temperature of the brass, t₂ = initial temperature of water, t₃ = temperature of the mixture.

Making C the subject of the equation

C = cm(t₃-t₂)/M(t₁-t₃)............................... Equation 2

Given: M = 0.59 kg, m = 2.8 kg, t₁ = 98 °C, t₂ = 5.0 °C, t₃ = 6.8 °C

Constant: c = 4200 J/kg.°C

Substitute into equation 2,

C = 2.8×4200(6.8-5.0)/0.59(98-6.8)

C = 21168/53.808

C = 393.399 J/kg.°C

Thus the specific heat capacity of the brass = 393.399 J/kg.°C

zhannawk [14.2K]2 years ago
3 0
The specific heat capacity of brass would be ranked between 0 and infinity
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The distance an object falls from rest through gravity is 
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           Distance  =  (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected 
if  ' D ' doesn't change but ' g ' does. 
So I'm going to start by rearranging the equation
to solve for ' t '.                                                      D  =  (1/2) (g) (t²)

Multiply each side by  2 :         2 D  =            g    t²  

Divide each side by ' g ' :      2 D/g =                  t² 

Square root each side:        t = √ (2D/g)

Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:

  -- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'

                                             and smaller 'g' ==> longer 't' .-- 

They don't change by the same factor, because  1/g  is inside the square root.  So 't' changes the same amount as  √1/g  does.

Gravity on the surface of the moon is roughly  1/6  the value of gravity on the surface of the Earth.

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5 0
2 years ago
A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio
dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

  • It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
  • In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
  • The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of  the mass of the child times the square of the distance to the center.
  • When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

       I_{to} = I_{d} + m*r^{2}  = m*\frac{r^{2}}{2} +  m*r^{2} = \frac{3}{2}*  m*r^{2} (1)

  • So, τ = 3/2*m*r²*α (2)
  • When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

       I_{t} = I_{d} + m*\frac{r^{2}}{4}   = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4}  = \frac{3}{4}*  m*r^{2} (3)

  • Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

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        same result than in (2), so the torque remains the same.

7 0
2 years ago
A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
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> Find the friction force on the crate

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Fh = Fph + Fgh 
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> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

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<span>= 0.47</span>

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raketka [301]

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Maximum negative distance = (-) (3)

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7 0
2 years ago
What type of roadway has the highest number of hazards per mile?
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