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Dmitry [639]
3 years ago
10

Some towns have decided to bury their garbage to get rid of it. What is one harmful effect of this practice?

Physics
2 answers:
arlik [135]3 years ago
4 0

Answer:

Decaying waste releases methane gas, a greenhouse gas

Explanation:

common sense that isn't very common

Lady bird [3.3K]3 years ago
4 0
Decaying waste releases methane gas,a greenhouse gas
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Tire or false
spayn [35]
1 and 4 are tire.
2 and 3 are not.
8 0
3 years ago
In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca
gizmo_the_mogwai [7]

Answer:

A) He finds the same value of q / m for different materials , B)      y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D)   B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

Therefore the third statement is correct

Explanation:

Part A

Thomson's experiments are the first proof that the atoms that until now were considered indivisible were constituted by different elements, in these experiments Thomson himself the ratio q / m of several cathodes and always found the same value, which allowed to establish that In atoms there are two types of particles, some of which are mobile and others are still.

When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

        F = ma

        q E = m a

        a = q / m E

We use the kinematic relationship

          y = voy t - ½ to t²

          x = v₀ₓ t

The initial vertical velocity of electrons is zero

           y = ½ a (x / v₀ₓ)²

We replace

           y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

         Fm = Fe

         q v B = q E

          v = E / B

Part D

       

        We replace

        y = ½ (q / m) E L² / (E / B)²

         y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

      -4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

       -16.97 10⁻⁴ = 6.54 10⁻³ B² q / m

      B² = -2.59 10⁻¹ q / m

      q / m = -1.758 10¹¹ C/ kg

      B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

      B = 2.13 10⁻⁶ T

Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct

8 0
3 years ago
A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude
finlep [7]
In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
                   = (5 * 10) Kg m/s
                   = 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
5 0
3 years ago
Of the solar radiation entering the atmosphere, about ___% reaches the surface directly as direct radiation.
strojnjashka [21]
Of the solar radiation entering the atmosphere, about 22.5% reaches the surface directly as direct radiation and is being absorbed. Thirty five percent of it is reflected back to the space. While 10.5% and 14.5% are scattered to the earth from the blue sky and from clouds, respectively. Also, 17.5 percent was being absorbed by the atmosphere. Thus leaving only 22.5% to be given directly to the Earth's surface. Also, another factor of this low value would be the transparency of the atmosphere. The atmosphere of the Earth has only an effective transparent to radiation from the sun of about .34 to .7 micrometer. 
8 0
3 years ago
When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If
masya89 [10]

The concept required to solve this problem is quantization of charge.

First the number of electrons will be calculated and then the total mass of the charge.

With these data it will be possible to calculate the percentage of load in the mass.

Q= ne

Here Q is the charge, n is the number of electrons and e is the charge on the electron

n = \frac{Q}{e}

Replacing,

n = \frac{4*10^{-6}C}{1.6*10^{-19}}

n = 2.5 * 10^{13}77

According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron

The total mass of the charge is

m= nm_e

Here,

m = Mass of the charge

n = Number of electrons

m_e = Mass of the electron

\text{Percentage change} = \frac{nm_e}{M}*100

Replacing we have

\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100

\text{Percentage change} = 6.9*10^{-14} \%

6 0
4 years ago
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