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natulia [17]
3 years ago
12

Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a to

tal of 11 rubs a distance of 7.50 cm each and with a frictional force averaging 71.3 N, what is the temperature increase? The mass of tissue warmed is only 0.100 kg, mostly in the palms and fingers. The specific heat of the tissue is 3500 J/(kg · °C).
Physics
1 answer:
stepan [7]3 years ago
6 0

Answer:

0.168°C

Explanation:

The frictional force between the palms enables rubbing and initiates the heat deposed.

The energy generated per rub is;

The force × distance

Hence we have; 71.3 N×0.0750m

Hence for 11 rubs we have; 11× 71.3 N×0.0750m= 58.82J

Now this heat acquired by the tissue is the same as mass of tissue × specific heat capacity of tissue × temperature rise;

Expressed mathematically as;

58.82=0.100 kg×3500 J/(kg · °C)×∆T

∆T= 58.82/ 0.100 ×3500 =0.168°C

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DIA [1.3K]

Answer:

74^{\circ} C

Explanation:

We are given that

Mass of glass,m=300 g

T_1=23^{\circ}

Volume,V=236cm^3

Mass of water=density\times volume=1\times 236=236 g

Density of water=1g/cm^3

Temperature of hot water,T=87^{\circ}

Specific heat of glass,C_g=0.2cal/g^{\circ}C

Specific heat of water,C_w=1 cal/g^{\circ}C

Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)

Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)

Q_{glass}+Q_{water}=0

300\times 0.2(T_f-23)+236\times 1(T_f-87)

60T_f-1380+236T_f-20532=0

296T_f=20532+1380=21912

T_f=\frac{21912}{296}=74^{\circ} C

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3 years ago
Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express
AlladinOne [14]

Answer:

Vb = k Q / r        r <R

Vb = k q / R³ (R² - r²)    r >R

Explanation:

The electic potential is defined by

             ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

             VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

           Va = - ∫ (k Q / r²) dr

           -Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

          Vb = k Q / r        r <R

         

We perform the calculation of the power with the expression of the electric field that they give us

           Vb = - int (kQ / R3 r) dr

  We integrate and evaluate from the starting point r = R to the final point r <R

         Vb = ∫kq / R³ r dr

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This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

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3 years ago
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mamaluj [8]

Answer:

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Answer:

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Hey

I have no idea when YOUR assignment is due.

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