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3 years ago

If a plant grows among many tall trees in a dense forest,how could it adapt so that it could get enough air and water

Physics

1 answer:

Nadusha1986 [10]3 years ago

4 0

Answer:

The answer is below

Explanation:

Plants that grow among many tall trees in a dense forest could adapt so that they could get enough air and water through the following means:

1. By becoming climbers. Here they climb up the nearby trees to get more air and sunlight.

2. Plants grow leaves that have pointed tips. This allows the passage of water in the plant's leaves without causing damage.

3. Some plants grow as epiphytes, where they grow on big trees' canopy

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An attempt to redirect internet traffic from a legitimate site to a different identical-looking site is categorized as which soc

Snezhnost [94]

Answer:PHARMING

Explanation:

Social engineering can be defined as an attempt to manage,change and regulate the future development and the behaviour of a society.

Common types of social engineering attack include; Watering hole, Pretexting, Phishing,Whaling attack, Baiting, PHARMING, and so on.

PHARMING is an attempt to redirect internet traffic from a legitimate site to a different identical-looking site. PHARMING is a spamming practice, a cyberattack and also, a type of phishing in which hackers use in stealing personal information from people on the internet.

PHARMING is done through the injection of malicious data or code into the victims' computer system. This injection is known as the DNS cache poisoning.

5 0

3 years ago

In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how t

kakasveta [241]

Answer:

N = 1036 times

Explanation:

The radial probability density of the hydrogen ground state is given by:

$p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }$

$p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }$

$p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }$

$N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}$

$N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } } }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}$

$N = 1300(16) e^{-3}$

N = 1035.57

N = 1036 times

6 0

3 years ago

While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?

sergiy2304 [10]

Answer:

option C

Explanation:

The correct answer is option C

When the driver takes the sharp right turn the door will exert rightward pressure on the driver.

When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.

The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.

3 0

3 years ago

9,10,and11 please I want the answer I can't found it

Ede4ka [16]

Q9. Inertia is the tendency of an object to resist any change in its motion.

Q10. A moving object's momentum depends on the object's mass and velocity.

Q11. I believe it would be air resistance. The fluid friction that opposes the motion of object's through the air.

7 0

3 years ago

1. A 3.5 kg object experiences an acceleration of 0.5 m/s2. What net force does the object experience

Leviafan [203]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.5 × 0.5 = 1.75

We have the final answer as

Hope this helps you

7 0

2 years ago