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Vedmedyk [2.9K]

3 years ago

7

Ultraviolet rays can cause skin cancer in humans and other living things. Explain why the depletion of the ozone layer by pollut

ants has increased the risk of skin cancer?

2 answers:

Charra [1.4K]3 years ago

6 0

Teh humans are more revealed to the sun and that causes cancer because of all the uv rays and the ozone is the only thing thats helping us from uv rays

user100 [1]3 years ago

6 0

The ozone layer blocks out most ultraviolet rays, which long exposure to can cause skin cancer. Pollutants deplete the ozone layer, making it weaker and putting holes in it. Ultraviolet rays can easily get through these holes at a higher concentration, therefore effecting people more.

You might be interested in

What are possible units for impulse? Check all that apply.

Neporo4naja [7]

Units of impulse: N • s, kg • meters per second

Explanation:

Impulse is defined in two ways:

1)

Impulse is defined as the product between the force exerted in a collision and the duration of the collision:

$I=F\Delta t$

where

F is the force

$\Delta t$ is the time interval

Since the force is measured in Newtons (N) and the time is measured in seconds (s), the units for the impulse are

$[I] = [N][s]$

So,

N • s

2)

Impulse is also defined as the change in momentum experienced by an object:

$I=\Delta p$

where the change in momentum is given by

$\Delta p = m\Delta v$

where m is the mass and $\Delta v$ is the change in velocity.

The mass is measured in kilograms (kg) while the change in velocity is measured in metres per second (m/s), therefore the units for impulse are

$[I]=[kg][m/s]$

so,

kg • meters per second

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

6 0

3 years ago

Read 2 more answers

The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensit

sveticcg [70]

Answer:

a) I = 3.63 W / m² , b) I = 0.750 W / m²

Explanation:

The intensity of a sound wave is given by the relation

I = P / A = ½ ρ v (2π f $s_{max}$ )²

I = (½ ρ v 4π² s_{max}²) f²

a) with the initial condition let's call the intensity Io

cte = (½ ρ v 4π² s_{max}²)

I₀ = cte s² f₀²

I₀ = cte 10 6

If frequency is increase f = 2.20 10³ Hz

I = constant (2.20 10³) 2

I = cte 4.84 10⁶

let's find the relationship of the two quantities

I / Io = 4.84

I = 4.84 Io

I = 4.84 0.750

I = 3.63 W / m²

b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or

I = cte (f s)²

I = constant (0.250 10³ 4)²

I = cte 1 10⁶

the relationship

I / Io = 1

I = Io

I = 0.750 W / m²

6 0

2 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0

3 years ago

What is a prediction

faust18 [17]

An educated guess about something. (What might happen in the future)

7 0

3 years ago

Read 2 more answers

Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i

Arte-miy333 [17]

I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
$R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i$
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
$R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_x=0.96m$
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
$R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_y=-0.84m$
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
$v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}$
$v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}$
Total velocity is:
$v=\sqrt{v_x^2+v_y^2}$
Once we calculate velocities we get:
$v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}$
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

3 0

3 years ago