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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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3 0
3 years ago
A shopper walks westward 2.8 meters in the first 5 minutes and then eastward 9.2 meters in the next 10 minutes. What is the shop
Alecsey [184]

Answer:

12m

Explanation:

Given parameters:

Distance walked westward  = 2.8m

Time of travel  = 5min

Distance walked eastward = 9.2m

Time of travel  = 10min

Unknown:

The total shopper's travel distance  = ?

Solution:

Total distance traveled is the sum of the length of path covered by a body. It is a scalar quantity.

 Total distance = distance walked westward + distance walked eastward

  Total distance  = 2.8m + 9.2m  = 12m

6 0
2 years ago
When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point
vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity \rightarrow 20m/s

\lambda = Wavelength \rightarrow 35*10^{-2}m

Therefore the frequency would be given as

f = \frac{20}{35*10^{-2}}

f = 57.14Hz

The frequency is directly proportional to the angular velocity therefore

\omega = 2\pi f

\omega = 2\pi *57.14

\omega = 359.03rad/s

Now the maximum speed from the simple harmonic movement is given by

V_{max} = A\omega

Where

A = Amplitude

Then replacing,

V_{max} = (1*10^{-2})(359.03)

V_{max} = 3.59m/s

Therefore the maximum speed of a point on the string is 3.59m/s

8 0
3 years ago
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