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3 years ago

Higher temperatures along with moisture climates generally increase the rate of

Physics

1 answer:

Bad White [126]3 years ago

4 0

Answer:

Option (D)

Explanation:

Weathering refers to the disintegration of rocks due to the different geological processes that are initiated by agents such as wind, water and ice.

In a region where the temperature is high, and there is sufficient amount of moisture content in the air, there occurs high rate of physical, chemical and biological type of weathering.

Physical weathering processes such as freeze-thaw method and exfoliation method commonly occurs in this type of region.
Chemical weathering processes such as hydrolysis, oxidation, as well as carbonation method occurs in the region of high temperature, moist climatic condition.
Biological weathering also is possible up to some extent as the plants takes up water from the rainfall and also receive enough sunlight for its growth, as a result of which it leads to the breakdown of rocks.

Thus, the correct answer is option (D).

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A car starts at the top of a hill with 200 J of energy, and rolls down a frictionless surface. The isolated system consists of t

Kamila [148]

Answer:

The total energy stays the same but is converted from being stored as gravitational potential energy into kinetic energy of the car as it moves.

Explanation:

the law of conservation of energy states that the total energy of an isolated system remains constant, and since it is gaining speed that energy will be kinetic

6 0

2 years ago

State and prove bessel inequality

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

Read 2 more answers

Sound, light, feelings and ideas are not

taurus [48]

I think it’s A, I’m so sorry if I’m wrong.

6 0

2 years ago

A motorcycle accelerates uniformly from rest at 7.9\,\dfrac{\text{m}}{\text{s}^2}7.9 s 2 m 7, point, 9, space, start fraction,

8090 [49]

Answer:

t = 3.516 s

Explanation:

The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

$v(t) = v_0 +at$

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0

If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

1 km = 1000 m

1 h = 3600 s

Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

100 km/h = 27.777... m/s

Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:

27.777 m/s = 7.9 m/s^2 t

Solving for t

t = (27.7777 / 7.9) s = 3.516 s

6 0

3 years ago

A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the

andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

$F = mg$

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

$\sum \tau = 0$

Regarding the forces we have,

$3Mg-1mg=0$

Re-arrange to find M,

$M = \frac{m}{3}$

$M = \frac{50}{3}$

$M = 16.67Kg$

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0

2 years ago