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laiz [17]
3 years ago
8

Plants in the desert, such as the yucca tree below, tend to have very wide root structures. Some yucca tree roots have been know

n to stretch more than 10 meters away from the trunk. The plant does this so that it can have access to as much of the water that falls in that area as possible. Which of the following best explains why yucca trees rarely grow close to other large plants in the desert? A. They compete with other large plants for sunlight. B. They are able to share water with nearby plants. C. They compete with other large plants for land. D. Staying far away from other large plants keeps them cool.
Physics
2 answers:
serious [3.7K]3 years ago
5 0
C. They compete with other large plants for land. Because if they have very wide roots they will take up more land so if another one is right next to the yuca tree they might both die or one will die from lack of water. Makes sense?
Pepsi [2]3 years ago
5 0
They compete with other large plants for land (study island)
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Every sample of a pure substance has exactly the same composition and the same properties. true or false
kirill115 [55]
False, that does not apply to some

7 0
2 years ago
How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?
aalyn [17]
There is too much information given, it's hard to understand exactly which variables are important in this problem. 
7 0
2 years ago
(d) What type of transformation is done by the following things?
mafiozo [28]

Answer:

human body is answer according to our studies

7 0
2 years ago
A bar on a hinge starts from rest and rotates with an angular acceleration α = 12 + 5t, where α is in rad/s2 and t is in seconds
Rama09 [41]

Answer:

Ф = 239.73 rad

Explanation:

α = 12 + 15×t

W = ∫α×dt

   = ∫(12 + 5×t)×dt

   = 12×t + 2.5×t^2

then:

Ф = ∫W×dt

   = ∫(12×t + 2.5×t^2)dt

   = 6×t^2 + 5/6×t^3

therefore the angle at t = 4.88s is:

Ф = 6×(4.88)^2 + 5/6×(4.88)^3

   = 239.73 rad    

5 0
3 years ago
HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0
3 years ago
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