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klemol [59]
3 years ago
10

. 2. El calor específico de un líquido es de 4186 j/kg.K. Su masa es de 100g. ¿Qué cantidad de calor hay

Physics
1 answer:
mote1985 [20]3 years ago
4 0

Answer:

Calor absorbido, Q = 5023.2 Joules

Explanation:

Dados los siguientes datos;

Masa = 100g to kg = 100/1000 = 0.1kg

Temperatura inicial, T1 = 293.15 K

Temperatura final, T2 = 305.15K

Capacidad calorífica específica = 4186 j/kg.K.

Para encontrar la cantidad de calor necesaria;

La capacidad calorífica viene dada por la fórmula;

Q = mct

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 305.15 - 293.15

dt = 12 K

Sustituyendo en la fórmula, tenemos;

Q = 0.1*4186*12

Q = 5023.2

Calor absorbido, Q = 5023.2 Joules

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The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

p_i = (M+m)u

where

M = 61 kg is the mass of the skater

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After, the ball is thrown at twice the velocity, so the final total momentum is

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3 years ago
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f
Hitman42 [59]
(a) The angular acceleration of the wheel is given by
\alpha =  \frac{\omega_f - \omega_i }{t}
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In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
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(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
\theta (t) = \omega_i t +  \frac{1}{2} \alpha t^2
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Two technicians are discussing fuse testing. Technician A says that a test light should light on both test points of the fuse if
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Answer:

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