Identify the physical property of a material that is NOT a good conductor of heat.

In order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the e

Mariulka [41]

Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

F = m a

G m M / x² = m dv / dt = m dv/dx dx/dt

G M / x² = dv/dx v

GM dx / x² = v dv

We integrate

v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m and the velocity v = vo and the upper limit x = 2.50 10⁸m with a velocity of v = 8.50 10³ m/s

½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3 10⁻⁸)

72.25 10⁶ - v₀² = -1.213 10⁸

v₀² = 72.25 10⁶ + 1,213 10⁸

v₀² = 193.6 10⁶

v₀ = 13.9 10³ m / s

6 0

3 years ago

If The density of this stainless steel is7.85 g/cm3,specific heatis 0.5 J/g.K, melting pointis 1673K, heat of fusion s0.260J/kg.

kodGreya [7K]

Answer:

$\Delta H=687.4 J$

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

$\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J$

$\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J$

Therefore, the required heat is:

$\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J$

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

4 0

3 years ago

Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet

yaroslaw [1]

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

Mass flow rate = 2 kg/s

Diameter of inlet pipe = 5.2 cm