Which of the following is an ecosystem

Four students are each pushing on one side of a box. Mara is pushing to the west, Cindi is pushing to the east, Chris is pushing

jenyasd209 [6]

Answer:

cindi

Explanation:

cindi's work done is larger than all the other students combined

5 0

2 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration ( $a_c$ ) and the tangential acceleration ( $a_t$ ):

$a=\sqrt{a_c^2+a_t^2}$

We know that:

a = 4.4g

$a_c = 3.2 g$

So, we can find the tangential acceleration:

$a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2$

The angular acceleration is related to the tangential acceleration by

$\alpha = \frac{a_t}{r}$

where r = 10.9 m is the length of the centrifuge. Substituting,

$\alpha = \frac{29.6}{10.9}=2.7 rad/s^2$

Learn more about centripetal and angular acceleration here:

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8 0

3 years ago

Electromagnet Fluctation

Tems11 [23]

Answer:

answer choice B

Explanation:

6 0

3 years ago

A physics student skis down a hill, accelerating at a constant

ikadub [295]

<h3>Answer:</h3>

225 meters

<h3>Explanation:</h3>

Acceleration is the rate of change in velocity of an object in motion.

In our case we are given;

Acceleration, a = 2.0 m/s²

Time, t = 15 s

We are required to find the length of the slope;

Assuming the student started at rest, then the initial velocity, V₀ is Zero.

<h3>Step 1: Calculate the final velocity, Vf</h3>

Using the equation of linear motion;

Vf = V₀ + at

Therefore;

Vf = 0 + (2 × 15)

= 30 m/s

Thus, the final velocity of the student is 30 m/s

<h3>Step 2: Calculate the length (displacement) of the slope </h3>

Using the other equation of linear motion;

S = 0.5 at + V₀t

We can calculate the length, S of the slope

That is;

S = (0.5 × 2 × 15² ) - (0 × 15)

= 225 m

Therefore, the length of the slope is 225 m

6 0

3 years ago

A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of

lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0

2 years ago

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