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3 years ago

Climate Change has been a hot topic in the news, politics, and science. To help clarify the issue, address the following:

1 answer:

3 0

I can answer number 2 for you only. I don’t know if it’s correct or not this is what I THINK it is:

2. Burning fossil fuels creates greenhouse gases, which changes the climate.
Ps. Climate is long term weather. Aka weather over a long period of time.

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If you run with an average speed of 3.9 meters per second for 1,200 seconds, then what distance will you have traveled?

Rufina [12.5K]

Answer: 3.9 x 1200 = 4680

Explanation:

7 0

3 years ago

A rod 14.0 cm long is uniformly charged and has a total charge of -22.0 μC. Determine the magnitude and direction of the net ele

lesya692 [45]

Explanation:

It is given that,

Length of the rod, l = 14 cm = 0.14 m

Charge on the rod, $q=-22\ \mu C=-22\times 10^{-6}\ C$

We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m

Electric field at the axis of the rod is given by :

$E=\dfrac{\lambda}{2\pi \epsilon_o z}$

Where

$\lambda$ is the linear charge density of the rod,

$\lambda=\dfrac{q}{l}=\dfrac{-22\times 10^{-6}\ C}{0.14\ m}=-0.00015\ C/m$

$E=\dfrac{-0.00015}{2\pi\times 8.85\times 10^{-12}\times 0.36}$

E = -7493170.57 N/C

$E=-7.49\times 10^6\ N/C$

Negative sign shows that the electric field is acting in inwards direction. Hence, this is the required solution.

4 0

3 years ago

Please help me!

sergeinik [125]

C. The Densities are equal.

<h3>What is density?</h3>

Density is mass per unit volume or mass of a unit volume of a material substance.

If m1, V1 and D1 = mass, volume and density respectively of ball C

m2, V2 and D2 = mass, volume and density respectively of ball D

According to the Question ,

$V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}$

Therefore,

$\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1$

Hence, D1 = D2

Learn more about density here:brainly.com/question/15164682

#SPJ1

6 0

3 years ago

Two semiconductors are identical except that one has a band gap of 1.2 eV, while the other has a band gap of 1.1 eV. The room te

solong [7]

To solve this problem it is necessary to apply the relationship given by the intrinsic carrier concentration, in each of the phases.

The intrinsic carrier concentration is the number of electrons in the conduction band or the number of holes in the valence band in intrinsic material. This number of carriers depends on the band gap of the material and on the temperature of the material.

In general, this can be written mathematically as

$\eta_i = \sqrt{N_cN_v}e^{-\frac{E_g}{2KT}}$

Both are identical semiconductor but the difference is band gap which is:

$E_{g1} = 1.1eV$

$n_{i1} = 1*10^{19}m^{-3}$

$E_{g2} = 1.2eV$

$T=300K$

The ratio between the two phases are given as:

$\frac{\eta_{i1}}{\eta_{i2}} = \frac{e^{-\frac{E_{g1}}{2KT}}}{e^{-\frac{E_{g2}}{2KT}}}$

$\frac{\eta_{i1}}{\eta_{i2}} = e^{\frac{E_{g2}-E_{g1}}{2KT}}$

$\frac{\eta_{i1}}{\eta_{i2}} =e^{\frac{(1.2-1.1)(1.6*10^{-19})}{2(1.38*10^{-23})(300)}}$

$\frac{\eta_{i1}}{\eta_{i2}} =e^{-1.932367}$

$\frac{\eta_{i1}}{\eta_{i2}} =0.145$

Therefore the ratio of intrinsic carrier densities for the two materials at room temperature is 0.145

7 0

4 years ago

A nonconducting ring with a radius of 11.5 cm is uniformly charged with a total positive charge of 10.0 µC. The ring rotates at

zhuklara [117]

Answer:

$B=1.21*10^{-10}T$

Explanation:

The magnitude of the magnetic field on the axis of the ring is given by:

$B=\frac{\mu_0 IR^2}{2(r^2+R^2)^{\frac{3}{2}}}(1)$

$\mu_0$ is the permeability of free space, $I$ is the flowing current through the ring, $R$ is the ring's radius and $r$ is the distance to the center of the ring.

The flowing current through the ring is defined as the ring's charge divided into the time taken by the charge to complete one revolution, that is, the period $T=\frac{2\pi}{\omega}$ . So, we have:

$I=\frac{q}{T}\\I=\frac{q}{\frac{2\pi}{\omega}}\\\\I=\frac{\omega q}{2\pi}\\I=\frac{18\frac{rad}{s}(10*10^{-6}C)}{2\pi}\\I=2.87*10^{-5}A$

Now, replacing in (1):

$B=\frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(2.87*10^{-5}A)(0.115m)^2}{2((0.05m)^2+(0.115m)^2)^{\frac{3}{2}}}\\B=1.21*10^{-10}T$

6 0

4 years ago