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3 years ago

What kind of movement does the heat within the core and mantle directly generate?

1 answer:

7 0

Heat within the earth generates convection currents in the mantle. Convection currents represent the circular motion of molten rocks in the mantle driven by heat from the core. This circular motion is what drives the plates and creates the theory of plate tectonics which states that the earth crust is divided into plates that float on the semi-molten mantle below.

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A ___ is a machine which contains a lever and wedge that combine to make work easier.

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4 years ago

A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin

gogolik [260]

Answer:

a) 23 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum p₀, can be written as follows:

$p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

The final momentum pf, can be written as follows:

$p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

$v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.

6 0

3 years ago

A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in

Gekata [30.6K]

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Learn more about heat energy here:

brainly.com/question/11961649

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3 years ago