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nata0808 [166]
3 years ago
6

How can humans activities affect the frequency and impact of natural disasters

Physics
2 answers:
denpristay [2]3 years ago
8 0
Forest fires can be an impact 

Mkey [24]3 years ago
5 0

Answer:

Human activities affect the frequency and impact of natural disasters because of urbanization, deforestation, waste disposal, resource extraction, man-made structures, and greenhouse gases.

Explanation:

You might be interested in
What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? Assume the resistivity of t
yawa3891 [41]

Given Information:

Diameter of spherical cell = 0.040 mm

thickness = L = 9 nm

Resistivity =  ρ = 3.6×10⁷ Ω⋅m

Dielectric constant = k = 9.0

Required Information:

time constant = τ = ?

Answer:

time constant = 2.87×10⁻³ seconds

Explanation:

The time constant is given by

τ = RC

Where R is the resistance and C is the capacitance.

We know that resistivity of of any material is given by

ρ = RA/L

R =  ρL/A

Where area of spherical cell is given by

A = 4πr²

A = 4π(d/2)²

A = 4π(0.040×10⁻³/2)²

A = 5.026×10⁻⁹ m²

The resistance becomes

R =  (3.6×10⁷*9×10⁻⁹)/5.026×10⁻⁹

R = 6.45×10⁷ Ω

The capacitance of the cell membrane is given by

C = kεoA/L

Where k = 9 is the dielectric constant and εo = 8.854×10⁻¹² F/m

C = (9*8.854×10⁻¹²*5.026×10⁻⁹)/9×10⁻⁹

C = 44.5 pF

C = 44.5×10⁻¹² F

Therefore, the time constant is

τ = RC

τ = 6.45×10⁷*44.5×10⁻¹²

τ = 2.87×10⁻³ seconds

6 0
3 years ago
Say you dropped a cannonball from the 17.0–meter mast of a ship sailing at 2.0 meters/second. How far from the base of the mast
Mice21 [21]

The correct choice is A. 0 meters.

If you simply dropped the cannonball and didn't throw it horizontally,

then it'll fall straight down the mast and land on the deck right next to

the mast.


While you were up there holding it, before you dropped it, the cannonball

was moving horizontally, at 2.0 meters/second, along with the rest of the

ship and everything else aboard. It continued doing that after it dropped,

and from the point of view (in the reference frame) of the mast and everyone

on the ship, it fell straight down, parallel to the mast.


Now that we have that question answered, we can proceed to the more-

important ones. I answered the easy one, but YOU'll have to answer these:


==> WHY did you climb the mast carrying a cannonball ?

Have you been drinking sea water or bad rum ?


==> WHY did you drop it, and never even yell "LOOK OUT BELOW !" ?


==> How many formerly-able-bodied souls were injured by the

plummeting cannonball ?


==> What did everybody ELSE yell after the impact ?


==> What did they do to you after they brought you down ?

3 0
3 years ago
Read 2 more answers
What is the speed of sound in air at 40°C?
Blababa [14]
343 meters per second Is the answer
4 0
3 years ago
Read 2 more answers
Which statement correctly compares ultraviolet light to visible light? Ultraviolet light has both a lower frequency and longer w
Effectus [21]

The correct statement is

Ultraviolet light has both a higher frequency and a higher radiant energy than visible light.

because ultraviolet light has wavelength smaller than the visible light hence has a greater frequency as compared to visible light. (frequency is inversely related to wavelength. hence smaller the wavelength , greater will be the frequency)

we also know that the radiant energy is directly proportional to the frequency. hence greater the frequency , greater will be the radiant energy.

Since the frequency is greater for ultraviolet light , it radiant energy is also greater


7 0
3 years ago
Read 2 more answers
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

5 0
3 years ago
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