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Lubov Fominskaja [6]
4 years ago
6

Why one part of the earth's surface is arid and dry and a nearby part is lush and wet is explained by the term?

Physics
1 answer:
Darina [25.2K]4 years ago
8 0
The process that explains why one part of the earth's surface is arid and dry and a nearby part is lush and wet is areal differentiation. It is<span> an approach to geography that shows </span>the dependence of the distribution of physical and human phenomena and the relation  to each other from the physical location. Areal integration on the other hand is the approach that studies how places interact with each other.
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How does the strong nuclear force compare with the electrostatic force in the nucleus of an atom?
sergeinik [125]

C. The strong nuclear force is only attractive and acts over shorter distances

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4 years ago
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In a nuclear reactor 1 g of mass is converted into energy. How much energy in Joules is produced?
KengaRu [80]

Answer:

3 x 10^5  J

Explanation:

mass of substance, m = 1 g = 0.001 kg

Velocity of light, c = 3 x 10^8 m/s

According to the Einstein mass energy equivalence, the energy associated with the mass is given by

E = m c^2

E = 0.001 x 3 x 10^8

E = 3 x 10^5  J

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4 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
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Answer:

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Explanation:

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E₁ = mechanical energy at initial state [J]

E_{1}=E_{pot}+E_{kin}+E_{elas}\\

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E₂ = mechanical energy at final state [J]

E_{2}=E_{kin}+E_{pot}

Now we can use the first statement to get the first equation:

E_{1}+W_{1-2}=E_{2}

where:

W₁₋₂ = work from the state 1 to 2.

E_{k}=\frac{1}{2} *m*v^{2} \\

E_{pot}=m*g*h

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

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58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]

4 0
3 years ago
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barxatty [35]
An ice cube would transfer heat to another object whose temperature
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A block of "dry ice" is sitting there at a temperature of -78°C (-109°F).
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If you could find a block of solid nitrogen, its temperature would be
63K (-210°C, -346°F).  An ice cube would transfer heat to that baby
so fast that it would instantly boil.

6 0
4 years ago
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