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2 years ago

What are two ways erosion affects California land

Physics

2 answers:

V125BC [204]2 years ago

7 0

Answer:

Water pollution and increased runoff

Explanation:

Erosion can cause soil infected with fertilisers, pesticides, ets. can travel to water bodies, causing that water to be polluted.

Erosion can send particles to block pores in the ground, "patching up" the ground, letting water stay on the ground instead of seeping underground, eventually leading to runoff.

Ymorist [56]2 years ago

6 0

Answer:

Two ways erosion affects California land are rising sea levels and soil erosion. Rising Sea Levels can cause homes to go under water and soil erosion can cause the ground to not be harvestable.

Explanation:

Erosions can have a huge effect on many things consisting in California.

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Which statements are true about moving the compass around the wire

GREYUIT [131]

Answer:

The statements are missing!

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3 years ago

Based on what you have learned about gravity, mass, and distance, write one paragraph to answer the following questions.

natka813 [3]

Bigger objects have more gravitational pull. Bigger, meaning the size and mass. The closer you are to the object, the stronger the pull is. The earth is much bigger, and has a bigger mass than the moon, meaning that the moon is able to orbit around the earth. You are closer to the earth, so you dont randomly get pulled towards thw moon, because earts gravitational pull is stronger than the moons.

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3 years ago

Calculate the electrostatic force that a small sphere A, possessing a net charge of 2.0 x 10^-6 Coulombs exerts on another small

Anni [7]

Answer:

F = 5.33*10^-4N

Explanation:

to find the electrostatic force you use the Coulomb's law, given by the formula:

$F=k\frac{q_Aq_B}{r^2}$

k: Coulomb's constant = 8.89*10^9 Nm^2/C^2

q_a: charge of A = 2.0*10^{-6}C

q_B: charge of B = -3.0*10^{-6}C

r: distance between the spheres = 10.0m

By replacing all these values you obtain:

$F=(8.89*10^9Nm^2/C^2)\frac{(2.0*10^{-6}C)(-3.0*10^{-6}C)}{(10.0m)^2}=5.33*10^{-4}N$

hence, the forcebetween the spheres is about 5.33*10^-4N

3 0

3 years ago

Block 2 of mass 1.00 kg is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 20

son4ous [18]

Answer:

x = 0.327 m

Explanation:

Block 2 of mass 1.00 kg is at rest

spring constant = 200 N/m

Block 1 of mass 2 kg moving at 4 m/s

m₁ v₁ = (m₁ + m₂)V

2 x 4 = (2 + 1) V

V = 2.67 m/s

loss of kinetic energy = gain elastic potential energy

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2$

$\dfrac{1}{2}\times (3)\times 2.67^2 = \dfrac{1}{2}\times 200 \times x^2$

x = 0.327 m

hence, the spring compressed distance is equal to x = 0.327 m

8 0

3 years ago

A frictionless piston–cylinder device contains 7 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin

lora16 [44]

Answer: -1038.8 kJ

Explanation:

From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.

from the question we were given

Mass of nitrogen, m = 7kg

initial temperature, T1 = 250k

Final temperature, T2 = 450k

Polytropic index, n = 1.4

Specific gas constant, R = 0.2968kJ/kgK

W = [p2 * v2 - p1 * v1] / 1 - n

W = [m * R * T2 - T1] / 1 - n

W = 7*0.2968*(450 - 250)] / 1 - 1.4

W = [7*0.2968*200] / -0.4

W = 415.52 / -0.4

W = -1038.8 kJ

5 0

3 years ago