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Dennis_Churaev [7]
3 years ago
5

A 3.5 kg gold bar at 94°C is dropped into 0.20 kg of water at 22°C What is the final temperature? Assume the specific heat of go

ld is 129 J/kg C Answer in units of °C
Physics
1 answer:
kramer3 years ago
8 0

Answer:

47.17 degree C

Explanation:

mg = 3.5 kg, T1 = 94 degree C, sg = 129 J/kg C

mw = 0.2 kg, T2 = 22 degree C, sw = 4200 J/kg C

Let T be the temperature at equilibrium.

Heat given by the gold = Heat taken by water

mg x sg x (T1 - T) = mw x sw x (T - T2)

3.5 x 129 x (94 - T) = 0.2 x 4200 x (T - 22)

42441 - 451.5 T = 840 T - 18480

60921 = 1291.5 T

T = 47.17 degree C

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Answer:

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I don’t know if these are correct please help <br> Will mark brainliest :)
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a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec
MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.  

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

\frac{1}{2}  mu_1^2+0=\frac{1}{2}  mv_1^2+\frac{1}{2}  mv_2^2

∴  \frac{1}{2}  m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2

∴ \frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2}  mv_1^2+\frac{1}{2} mv_2^2

∴ mv₁v₂ = 0

  1. It is impossible for the mass to be zero.
  2. Because the second ball moves, velocity v2 cannot be zero.
  3. As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

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2 years ago
A 20×10⁹charge is moved between two points A andB that are 30mm apart and have an electric potential difference of 600v between
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Answer:

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Explanation:

7 0
3 years ago
How is this formula ( Rt= r1 x r2 / r1 + r2) which is used to calculate total resistance of two parallel resistors found from th
Maru [420]

Explanation:

The total resistance used to calculate the total resistance of the two parallel resistor is given by :

R_t=\dfrac{r_1\times r_2}{r_1+r_2}

Taking reciprocal of the above equation.

\dfrac{1}{R_t}=\dfrac{r_1+r_2}{r_1\times r_2}

We can also write the above equation as :

\dfrac{1}{R_t}=\dfrac{r_1}{r_1\times r_2}+\dfrac{r_2}{r_1\times r_2}

On simplification of above equation,

\dfrac{1}{R_t}=\dfrac{1}{r_2}+\dfrac{1}{r_1}

or

\dfrac{1}{R_t}=\dfrac{1}{r_1}+\dfrac{1}{r_2}

Hence, proved.

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