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STatiana [176]
3 years ago
13

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression

in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.
Physics
1 answer:
levacccp [35]3 years ago
6 0

maximum speed of cheetah is

v_1 = v_{max}

speed of gazelle is given as

v_2 = v_{g}

Now the relative speed of Cheetah with respect to Gazelle

v_{12} = v_1 - v_2

v_{12} = v_{max} - v_g

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

d = v_{12}* t

so by rearranging the terms we can say

t = \frac{d}{v_{12}}

t = \frac{d}{v_{max} - v_g}

so above is the relation between all given variable

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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:

brainly.com/question/2562955

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

8 0
3 years ago
Yellow-green light has a wavelength of 560 nm. What is its frequency?
Natasha2012 [34]
5.4 x 1014Hz
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7 0
2 years ago
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"Two long parallel wires 24.0 cm apart carry currents of 3.0 A and 8.0 A in the same direction. How far from the wire carrying 3
avanturin [10]

Answer:

6.5454 m

Explanation:

Let the distance from the wire carrying 3 A current is x

Then the distance from the the carrying current 8 A is 24-x

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It is given that magnetic field is zero at some distance so

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Here i_1=3\ A \ and\  i_2=8\ A

So \frac{3}{x}=\frac{8}{24-x}=6.5454\ m

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alexandr1967 [171]
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Deffense [45]

Answer:

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-electromagnetic waves – eg light waves, microwaves, radio waves.

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Explanation:

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7 0
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