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3 years ago

12

How much has the Earth’s climate warmed over the last 100 years?

1 answer:

GenaCL600 [577]3 years ago

4 0

<span>The Earth’s climate warmed over the last 100 years beyond 1°, although it is not still arrived at 1.1°. Only one grade can be perceived as a small warm, but the repercussions are catastrophic, because of the inertia possessed by the physics mechanisms that are regulating our planet. In order to apply fully the Paris Agreement, we should keep the warm well under 2°, aiming to keep it under 1.5°.</span>

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Which of these is a benefit of social networking ?

Alisiya [41]

Answer:

Staying connected to friends

Explanation:

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3 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

3 years ago

A joule is an amount of energy, and a watt is a rate of using energy, defined as 1 W = 1 J / s. How many joules of energy are re

Naily [24]

How many joules of energy are required to run a 100 W light bulb for one day?

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3 years ago

What is the average velocity of the particle from rest to 9 seconds?

geniusboy [140]

Answer: v = 2[m/s]Explanation:This avarage velocity can be found with the ... B. 2 meters/ second. C. 3 meters/second. D. 4 meters/second. 1.

7 0

3 years ago

Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat

STALIN [3.7K]

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

$B = \frac{\mu_0}{2}\frac{Ia^2}{(z^2+a^2)^{3/2}}$

Where,

a = Radius

z = Distance to the magnetic field

I = Current

$\mu_0 =$ Permeability constant in free space

Our values are given as

$D=2a = 16cm \rightarrow$ diameter of the sphere then,

$a = 0.08m$

Thus z = a

$B = \frac{\mu_0}{2}\frac{Ia^2}{(a^2+a^2)^{3/2}}$

$B = \frac{\mu_0I}{2(2^{3/2})a}$

$B = \frac{\mu_0 I}{2^{5/2}a}$

Re-arrange to find I,

$I = \frac{2^{5/2}Ba}{\mu_0}$

$I = \frac{2^{5/2}(3*10^{-12})(8*10^{-2})}{4\pi*10^{-7}}$

$I = 1.08*10^{-6}A$

Therefore the current at the pole of this sphere is $1.08*10^{-6}A$

5 0

3 years ago