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joja [24]
3 years ago
10

Which of the following can reduce biodiversity within a natural environment? A. Yucca moths pollinate and feed on yucca plants i

n the Mojave Desert. B. Red-billed oxpeckers eat ticks off of impalas' coats in the grasslands of East Africa. C. Non-native American bullfrogs out-compete native amphibians in the Tropical Andes. D. Giant kangaroo rats enrich the soil for plants with their piles of grass clippings in the California grasslands.
Physics
1 answer:
DaniilM [7]3 years ago
8 0
I wanna say B Red-billed oxpeckers eat ticks off of impalas
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Suppose the block is released from rest with the spring compressed 5.00 cm. The mass of the block is 1.70 kg and the force const
IRISSAK [1]

First, let's calculate the total mechanical energy when the block is at rest and the spring is compressed 5 cm:

\begin{gathered} ME=PE+KE\\ \\ ME=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ ME=\frac{955\cdot0.05^2}{2}+0\\ \\ ME=1.194\text{ J} \end{gathered}

Now, let's use this total energy to calculate the velocity when the spring is compressed by 2.5 cm:

\begin{gathered} ME=PE+KE\\ \\ 1.194=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ 2.388=955\cdot0.025^2+1.7v^2\\ \\ 1.7v^2=2.388-0.597\\ \\ 1.7v^2=1.791\\ \\ v^2=\frac{1.791}{1.7}\\ \\ v^2=1.0535\\ \\ v=1.026\text{ m/s} \end{gathered}

Therefore the speed is 1.026 m/s.

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1 year ago
Under which of the following conditions is Lactic acid fermentation most likely occur?
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Lactic acid is caused by using atp without oxygen being avaliable
5 0
3 years ago
A 6 kg block is sliding down a horizontal frictionless surface with a constant speed of 5 m/s. It then slides down a frictionles
BARSIC [14]

Answer:

5in

Explanation:

4 0
3 years ago
A particle travels in a circle of radius 76 cm and completes one revolution in 4.5 s. What is the centripetal acceleration of th
Dimas [21]

r = radius of the circle traveled by the particle = 76 cm = 0.76 m

T = time period of revolution for the particle = 4.5 s

w = angular velocity of the particle

angular velocity of the particle is given as

w = 2π/T

inserting the values

w = 2 (3.14)/4.5

w = 1.4 rad/s

a = centripetal acceleration of the particle in the circle

centripetal acceleration is given as

a = r w²

inserting the values

a = (0.76) (1.4)²

a = 1.5 m/s²

3 0
3 years ago
Read 2 more answers
What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0
2 years ago
Read 2 more answers
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