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3 months ago

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Which process is a form of mechanical weathering?

2 answers:

Yuliya22 [3.3K]3 months ago

8 0

Mechanical weathering refers to the natural fragmentation of rocks into smaller fragments through physical processes. This type of weathering occurs without altering the chemical composition of the rocks. Examining the options: 1. Hydration involves a substance absorbing water. 2. Carbonation relates to the dissolving of carbon dioxide in liquid, predominantly water. 3. Oxidation is the interaction of oxygen with elements. 4. Exfoliation describes the process where rocks are worn away in layers rather than grain by grain. Clearly, the last option, Exfoliation, aligns with the definition of mechanical weathering, making it one of its forms.

ValentinkaMS [3.4K]3 months ago

7 0

thermal expansion

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An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.51 μC/m2. Another infini

Sav [3153]

E_total = 5.8 x 10⁴ N/C Explanation: To determine the electric field at specified points, we must calculate the vectors individually for each charge and sum them. The electric field caused by each charged conductive sheet can be derived via Gauss's law with the understanding of scalar products between the electric field and relevant surfaces.

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3 months ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

Softa [3030]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Charge (Q₁) = 572.8 $* 10^{-9}$ C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² = 0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

8 0

3 months ago

The filament of the light bulb is made of tungsten. The resistance of the light bulb at room temperature (20∘C), measured by an

Ostrovityanka [3204]

Explanation:

The formula illustrating the relationship between resistance and temperature is as follows:

R =

$R_{o} + \alpha [T_{2} - T_{1}]$

where, R = final resistance

= initial resistance $R_{o}$

= temperature coefficient of resistivity $\alpha$

= final temperature $T_{2}$

= initial temperature $T_{1}$

Given data as follows.

$T_{1} = (20 + 273) K = 293 K$

R = 36 ohm,

= 3 ohm $T_{2}$

= 0.0045 $R_{o}$

Substituting the provided values into the above formula gives us the following.

R = $\alpha$

36 =

$R_{o} + \alpha [T_{2} - T_{1}]$

=

$3 + 0.0045 \times [T_{2} - 293]$

= 7626.33 K

$T_{2}$ Thus, it can be concluded that $\frac{34.3185}{0.0045}$ the temperature of the light bulb at 12.0 V is 7626.33 K.

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3 months ago