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pishuonlain [190]
3 years ago
10

Explain what is the difference of physical and chemical weathering

Physics
1 answer:
shusha [124]3 years ago
5 0
The difference is that physical weathering is a process that weathers rock without a chemical reaction or change. Chemical weathering changes the identity of rocks and it involves a chemical reaction or change.
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The solar system is really big! To give you an idea of its dimensions, think that if the Sun was Washington DC and Mercury was t
12345 [234]

The sun is the center of the solar system which is surrounded by the nine planets and they tend to orbit the sun in concentric circles with the sun as the center.

<h3>What is the solar system?</h3>

The term solar system has to do with the arrangement of the sun and the planets. The sun lies at the focus of the solar system. Now we know that there are nine planets that orbit around the sun. The distance between the planets and the sun depends on their relative proximity to each other.

Thus, the sun is the center of the solar system which is surrounded by the nine planets and they tend to orbit the sun in concentric circles with the sun as the center.

Learn more about solar system:brainly.com/question/12075871

#SPJ1

8 0
1 year ago
A girl throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounce
Karolina [17]

Answer:

F = 352 N

Explanation:

we know that:

F*t = ΔP

so:

F*t = MV_f-MV_i

where F is the force excerted by the wall, t is the time, M the mass of the ball, V_f the final velocity of the ball and V_i the initial velocity.

Replacing values, we get:

F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)

solving for F:

F = 352 N

 

3 0
3 years ago
When do we land on Mars ?
lorasvet [3.4K]

Answer:

I think like 2024 or 25

Explanation:

Elon musk will probably go

5 0
3 years ago
Read 2 more answers
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride, F−. How many grams of
Leviafan [203]

Answer:

2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-

Explanation:

Here are the steps of how to arrive at the answer:

The volume of a cylinder = ((pi)D²/4) × H

Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m

H = Height of the reservoir = 87.32m

Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³

1ppm = 1g/m³

0.8ppm = 0.8 × 1g/m³

= 0.8g/m³

Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.

Thank you for reading.

4 0
3 years ago
A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut
blsea [12.9K]

Answer:

1.E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2.E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

\rho (r) = \alpha (1-\frac{r}{R})

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

\int\, da = 2\pi r h

where ‘h’ is the length of the imaginary Gaussian surface.

Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3. At the boundary where r = R:

E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}

As can be seen from above, two E-field values are equal as predicted.

4 0
3 years ago
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