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brilliants [131]
3 years ago
10

Can you rule out natural factors as the main cause of today's climate change?

Physics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

Yes, as today human activities are the main cause for climate change.

In recent years, due to the over-consumption of primary sector products, the reckless over-consumption of natural resources and the increase of the earth's population, the natural environment has deteriorated, resulting in an imbalance between the countries of the developed and developing world in terms of climatic and environmental security. The global purpose of states, institutions and associations is to work together for sustainable development at all geographical levels and to fight inequality at the international level. At the same time, it is characteristic that a global movement has developed to demand the equal distribution of environmental burdens and climate justice.  

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At an instant when the displacement is equal to a/2, what fraction of the total energy of the system is potential?
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4 0
3 years ago
Suppose the microphone on an AM radio is not working properly. Which best explains how this will affect the process of amplitude
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4 years ago
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Which of the following equations is balanced correctly? A. 3 H2O → H2 + 3 O2 B. Cl2 + 2 KBr → KCl + Br2 C. 2 C2H2 + 5 O2 → 4 CO2
posledela
A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced. That leaves C. to be correctly balanced.
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3 years ago
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Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
3 years ago
Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters
Vesnalui [34]

Answer:

C. \frac{3F}{8}

Explanation:

Let initial charges on both spheres be,q

F=\frac{Kq^2}{d^2}   \ \ \  \ \ \  \ \ \  \ \_i

When the sphere C is touched by A, the final charges on both will be,\frac{q}{2}

#Now, when C is touched by B, the final charges on both of them will be:

q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\

Now the force between A and B is calculated as:

F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}

Hence the electrostatic force becomes 3F/8

4 0
3 years ago
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