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2 years ago

Sunburn is caused by

Physics

2 answers:

Snowcat [4.5K]2 years ago

5 0

Answer:

too much exposure to the sun's rays

Stells [14]2 years ago

3 0

Exposure to the sun

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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 61 m/s^2. If he reaches the gr

ira [324]

Answer:

55.66 m

Explanation:

While falling by 50 m , initial velocity u = 0

final velocity = v , height h = 50 , acceleration g = 9.8

v² = u² + 2gh

= 0 + 2 x 9.8 x 50

v = 31.3 m /s

After that deceleration comes into effect

In this case final velocity v = 17 m/s

initial velocity u = 31.3 m/s

acceleration a = - 61 m/s²

distance traveled h = ?

v² = u² + 2gh

(17)² = (31.3)² - 2x 61xh

h = 690.69 / 2 x 61

= 5.66 m

Total height during which he was in air

= 50 + 5.66

= 55.66 m

3 0

2 years ago

In a physics lab, light with a wavelength of 530 nm travels in air from a laser to a photocell in a time of 16.7 ns . When a sla

Harlamova29_29 [7]

Answer:

$\lambda'=78.086\ nm$

Explanation:

Given:

wavelength of light in the air, $\lambda=530\times 10^{-9}\ m$

time taken to travel from the source to the photocell via air, $t=16.7\ s$

time taken to reach the photocell via air and glass slab, $t'=21.3\times 10^{-9}\ s$

thickness of the glass slab, $x=0.87\ m$

<u>Now we have the relation for time:</u>

$\rm time=\frac{distance}{speed}$

hence,

$t=\frac{d}{c}$

c= speed of light in air

$16.7\times 10^{-9}=\frac{d}{3\times 10^8}$

$d=16.7\times 10^{-9}\times 3\times 10^8$

$d=5.01\ m$

For the case when glass slab is inserted between the path of light:

$\frac{(d-x)}{c} +\frac{x}{v} =t'$ (since light travel with the speed c only in the air)

here:

v = speed of light in the glass

$\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}$

$v=4.42\times 10^7\ m.s^{-1}$

Using Snell's law:

$\frac{\lambda}{\lambda'} =\frac{c}{v}$

$\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}$

$\lambda'=78.086\ nm$

5 0

2 years ago

When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration d

Mice21 [21]

The gravitational acceleration at any distance r is given by

$g= \frac{GM}{r^2}$

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is $r_e=6.37 \cdot 10^6 m$ , so the meteoroid is located at a distance of:

$r=r_e+2.60 r_e =3.60 r_e = 2.29 \cdot 10^7 m$

And by substituting this value into the previous formula, we can find the value of g at that altitude:

$g= \frac{GM}{r^2} = \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(2.29 \cdot 10^7 m)^2} =0.75 m/s^2$

5 0

3 years ago

Read 2 more answers

Calculate the potentia energy of a car with a mass of 3800kg that is on a hill 110 meters above sea level? USE 10 instead of 9.8

Arada [10]

Potential energy (PE ) = m g h

Where:

m = mass = 3800 kg

g = acceleration due gravity = 10 m/s^2

h = heigth = 110 meters

Replacing:

PE = 3800 * 10 * 110 = 4,180,000 J

7 0

8 months ago

Which of these experiments would make use of qualitative data

7nadin3 [17]

It has to due with numbers so I would say the last one!

5 0

2 years ago

Read 2 more answers