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max2010maxim
1 month ago
15

Which process is a form of mechanical weathering?

Physics
2 answers:
Yuliya22 [3.3K]1 month 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]1 month ago
7 0
thermal expansion
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Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -2.0 µC; sphere B carries a charge of -6.0 µC;
inna [3103]
None of the provided options is correct. After contact, A becomes -4 µC, B remains 0 µC, and C ends with +4.0 µC. When spheres A and B touch, charges will redistribute to establish balance, resulting in A = -4 µC, B = -4 µC, C = +4.0 µC. After C and B are touched, both positive and negative charges neutralize each other, leaving A at -4 µC, B at 0 µC, and C at 0 µC.
5 0
1 month ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
Softa [3030]

Answer:

57.94°

Explanation:

We understand that the formula for flux is

\Phi =E\times S\times COS\Theta

where Ф represents flux

           E indicates electric field

           S denotes surface area

        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
2 months ago
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
kicyunya [3294]

Answer:

Induced EMF is 2 x 10⁻³ volts

Explanation:

B = strength of the magnetic field aligning with the loop's axis = 1 T

\frac{dA}{dt} = area change rate of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = the angle formed by the magnetic field and area vector = 0

E = the induced EMF across the loop

EMF can be calculated using the formula

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
1 month ago
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
ValentinkaMS [3465]

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C represents the charge's magnitude

and \Delta V = 6.0 kV = 6000 V signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

4 0
2 months ago
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
kicyunya [3294]

Response:

Clarification:

The force between two charges, q₁ and q₂ at a distance d is represented by the formula

F = k q₁ q₂ / d²

Here, the force between charge q₁ = -15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C with distance d = (1.66 - 1.24) = 0.42 mm

k = 1 / (4π x 8.85 x 10⁻¹²)

Substituting the values into the equation

F = 1 / (4π x 8.85 x 10⁻¹²) x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 35969.4 x 10⁻³ N.

For the force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C at a distance d = (1.24 - 0) = 1.24 mm.

Substituting the values into the expression

F = 1 / (4π x 8.85 x 10⁻¹²) x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 82729.6 x 10⁻³ N

Both forces direct towards the left (away from the origin, towards the negative x-axis)

Total force = 118699 x 10⁻³

= 118.7 N.

5 0
1 month ago
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