When a certain amount of energy Q is supplied/given off to/from a sample of substance with mass m, the temperature of the substance increases/decreases by an amount $\Delta T$ , according to the equation

$Q=mC_s \Delta T$

where

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

$\Delta T$ is the change in temperature of the substance

In this problem, we have:

m = 1 kg = 1000 g is the mass of the concrete slab

$Q = -12,000 J$ is the amount of energy lost by the slab

$\Delta T = 30-26= -4^{\circ}C$ is the change in temperature of the slab

Solving the equation for $C_s$ , we find the specific heat capacity of concrete:

$C_s = \frac{Q}{m \Delta T}=\frac{-12,000}{(1000)(-4)}=3.0 J/(g^{\circ}C)$

Learn more about specific heat capacity:

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3 0

3 years ago

The acceleration of a particle as it moves along a straight line is given by a (2t 1) m/s2, where t is in seconds. Suppose that

Cerrena [4.2K]

Answer:

v = 158 m / s

Explanation:

In this kinematic problem they give us the acceleration of the body, the position and speed of the initial moment, let's use the definition of acceleration

a = dv / dt

dv = a dt

We integrate

∫ d v = ∫ (2t + 1) dt

v = 2 (2t²) + 1t

We evaluate between the initial time t = 0 that has speed (vo) 8 m/s and the final time t with speed v

v -8 = 4t2 + t

We already have the velocity formula as a function of time, let's calculate for t = 6 s

v = 8 + 4 6² +6

v = 158 m / s

3 0

4 years ago