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3 years ago

7

PLEASE HELP While the Earth is revolving around the sun, less direct sunlight is reaching the Northern Hemisphere than the South

ern Hemisphere. What season is it in the Northern Hemisphere?(1 point)
winter

autumn

spring

summer

1 answer:

jekas [21]3 years ago

7 0

Answer:

Winter

Explanation:

Earth Rotates about an axis

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Volume flow rate $= 1.81 * 10^{-2}$ meter cube per second

Explanation:

As we know that the

Pressure at the two ends would be the same along with volume of flow.

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and

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Substituting the given values in above equation we get

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3 years ago

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3 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

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For solving this we use the kinematic equations:

In the first part we have:

$h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\$

and the final speed is:

$v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T$

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

$v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2$

The sum of these heights will give us the total height, which is known:

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This is the time that its needed in order for the rocket to reach the required altitude.

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3 years ago

A gyroscope flywheel of radius 3.21 cm is accelerated from rest at 13.2 rad/s2 until its angular speed is 2450 rev/min.

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Answer:

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Part c)

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Part a)

as we know that angular acceleration of the wheel is given as

$\alpha = 13.2 rad/s^2$

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R = 3.21 cm

so the tangential acceleration is given as

$a_t = R\alpha$

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Part b)

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$f = 2450 rev/min$

$f = \frac{2450}{60} = 40.8 rev/s$

now we know that

$\omega = 2\pi f = 2\pi(40.8) = 256.56 rad/s$

now radial acceleration is given as

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Part c)

total angular displacement of the point on rim is given as

$\Delta \theta = \omega_0 t + \frac{1}{2}\alpha t^2$

here we know that

$\omega = \omega_0 + \alpha t$

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$t = 19.4 s$

now angular displacement will be

$\Delta \theta = 0 + \frac{1}{2}(13.2)(19.4)^2$

$\Delta \theta = 2493.3 rad$

now the distance moved by the point on the rim is given as

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7 0

3 years ago