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Nuetrik [128]
3 years ago
6

What phytophotodermatitis caused by?

Physics
2 answers:
tatiyna3 years ago
8 0

Answer:

Phytophotodermatitis is induced by the action of long wavelength ultraviolet radiation (UVA) on a plant chemical called furocoumarins (psoralens) on the skin surface. Contact with the plant, fruit or vegetable may have been brief and unnoticed.

Explanation:

Paraphin [41]3 years ago
3 0

Phytophotodermatitis happens when certain plant chemicals cause the skin to become inflamed following exposure to sunlight.

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If i am changing velocity, i must also have _______ and a net _________
Alinara [238K]

if i am changing velocity, i must also have <u>acceleration</u> and a net <u>force</u>

<h2><u>Newton's</u><u> </u><u>first</u><u> </u><u>law</u><u> </u><u>of</u><u> </u><u>motio</u><u>n</u></h2>

  • Newton's first law of motion states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

According to Newton's first law of motion, without a force acting on an object, its velocity does not change. The net force acts on an object to change its velocity and cause acceleration.

Read more about velocity:

brainly.com/question/4931057

6 0
3 years ago
Hardness and density are physical properties.<br> a. True<br> b. False
antiseptic1488 [7]
a . true hardness and density are physical properties
7 0
3 years ago
Read 2 more answers
What measures the radiation given off by earth
gladu [14]
 i thinkits an instrument called seismograph. not sure
3 0
3 years ago
A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C
Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

d_{ab} = distance traveled from station A to station B

t_{ab} = time of travel between station A to station B

we know that

Time = \frac{distance}{speed}

t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}

Total distance traveled is given as

d = d_{ab} + d_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }

Given that :

d_{ab} = 4 d_{bc}

So

v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}

2)

v_{ab} = Speed of train from station A to station B = 80 kmh⁻¹

t_{ab} = time of travel between station A to station B

d_{ab} = distance traveled from station A to station B

we know that

distance = (speed) (time)

d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}

d_{bc} = distance traveled from station B to station C

v_{bc} = Speed of train from station B to station C = 60 kmh⁻¹

t_{bc} = time of travel for train from station B to station C

we know that

distance = (speed) (time)

d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}

Total distance traveled is given as

d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}

Total time of travel is given as

t = t_{ab} + t_{bc}

Average speed is given as

v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}

Given that :

t_{ab} = 3 t_{bc}

So

v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}

4 0
3 years ago
In a lab, four balls have the same velocities but different masses.
olya-2409 [2.1K]

Answer:

New Momentum of Ball B=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}

<u>Explanation:</u>

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2\frac{m}{s}

Momentum of Ball A=2.2\frac{k g m}{s}

Momentum of Ball B=4.4 \frac{k g m}{s}

Momentum of Ball C=11\frac{k g m}{s}

Momentum of Ball D=15\frac{k g m}{s}

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2\frac{m}{s}

Initial Momentum of Ball B=4.4\frac{k g m}{s}

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

                                            =3×2=6kg

Thus we get Mass of Ball B=6kg

According to the formula,  

Change in momentum of Ball B \Delta p=m \times \Delta v

Where \Delta p=change in momentum

          m=mass of the ball B

         \Delta v=change in velocity ball B

And \Delta v=v, since all balls, have same velocity

Thus the above equation, changes to

         \Delta p=m \times v

Substitute all the values in the above equation we get

         \Delta p=6 \times 2.2

                     =13.2 \frac{\mathrm{kgm}}{\mathrm{s}}  

Result:

 Thus the New Momentum of ball B=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}

3 0
3 years ago
Read 2 more answers
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