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gavmur [86]
3 years ago
10

Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene

450 g de agua a 23 ºC. Sr agita la mezcla y la temperatura se incrementa hasta 26 ºC. ¿Cuál es el calor especifico de la plata?
Physics
1 answer:
sdas [7]3 years ago
8 0

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

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Question 2 (10 points)
kenny6666 [7]

The force required to slow the truck was -5020 N

Explanation:

First of all, we find the acceleration of the truck, which is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time

For the truck in this problem,

v = 11.5 m/s

u = 21.9 m/s

t = 2.88 s

So the acceleration is

a=\frac{11.5-21.9}{2.88}=-3.6 m/s^2

where the negative sign means that this is a deceleration.

Now we can find the force exerted on the truck, which is given by Newton's second law:

F=ma

where

m = 1390 kg is the mass of the truck

a=-3.6 m/s^2 is the acceleration

And substituting,

F=(1390)(-3.6)=-5004 N

So the closest answer among the option is -5020 N.

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

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#LearnwithBrainly

8 0
2 years ago
A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre
Brut [27]

Answer:

The time taken is  t =  0.356 \ s

Explanation:

From the question we are told that

  The length of steel the wire is  l_1  = 31.0 \ m

   The  length of the  copper wire is  l_2  = 17.0 \ m

    The  diameter of the wire is  d =  1.00 \ m  =  1.0 *10^{-3} \ m

     The  tension is  T  =  122 \ N

     

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              t  =  t_s  +  t_c

Where  t_s is the time taken to transverse the steel wire which is mathematically represented as

         t_s  = l_1 *  [ \sqrt{ \frac{\rho * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_s is the density of steel with a value  \rho_s  =  8920 \ kg/m^3

   So

      t_s  = 31 *  [ \sqrt{ \frac{8920 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_s  = 0.235 \ s

 And

        t_c is the time taken to transverse the copper wire which is mathematically represented as

      t_c  = l_2 *  [ \sqrt{ \frac{\rho_c * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_c is the density of steel with a value  \rho_s  =  7860 \ kg/m^3

 So

      t_c  = 17 *  [ \sqrt{ \frac{7860 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_c  =0.121

So  

   t  = t_c  + t_s

    t =  0.121 + 0.235

    t =  0.356 \ s

4 0
3 years ago
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