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3 years ago

Complete the simulation for three different trails. Complete chart below, using the data for your simulation from each trail.

Physics

2 answers:

777dan777 [17]3 years ago

7 0

Where’s the chart for the question?

jok3333 [9.3K]3 years ago

3 0

Answer:

Are similar in pattern

Explanation:

edge 2021

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11. Two point charges, initially 1 cm apart, are moved to a distance of 3 cm apart. By what factor

vampirchik [111]

Answer:

Explanation:

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3 years ago

What happens if you add additional,solid NaCl after the maximum has been reached?

azamat

<span>it would bond to the phosphate

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3 years ago

Read 2 more answers

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:

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):

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

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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#LearnwithBrainly

8 0

3 years ago

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the

dolphi86 [110]

Answer:

Please find the answer in the explanation

Explanation:

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

The slope of any time graph can not give you distance or displacement except for position - time graph.

When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.

You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.

5 0

3 years ago

There is a species of bamboo that can grow 36 inches per day. If a plant grew at this rate and was measured at 40 inches initial

LekaFEV [45]

Answer:

It will take the plant $4\frac{4}{9}$ days or 4.44 days to grow to a height of 200 inches tall.

Explanation:

From the question, the rate at which the species of the bamboo tree grows is 36 inches per day.

To determine how long it would take a plant 40 inches tall initially to grow at this rate (that is, 36 inches per day) to a height of 200 inches.

This means we will calculate the number of days it will take the plant to grow additional 160 inches ( 200 inches - 40 inches) at this rate.

Now,

If the plant grows 36 inches in 1 day

then it will grow 160 inches in x days

x = (160 inches × 1 day) / 36 inches

x = 160 / 36

x = $4\frac{4}{9}$ days or 4.44 days

Hence, it will take the plant $4\frac{4}{9}$ days or 4.44 days to grow to a height of 200 inches tall.

8 0

3 years ago