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3 years ago

6

Wings of a bird what kind of motion is it

1 answer:

Lana71 [14]3 years ago

4 0

Answer:

linear motion

Explanation:

the birds in the sky show oscillatory motion when they flap their wings

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Tom rides his motorcycle at a speed of 15 meters/second for an hour.

Dahasolnce [82]

Answer:

1 hour to ride his motorcycle

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3 years ago

1. Explain one way gravity affects objects.

Alex Ar [27]

Answer:

it makes it fall to the ground instead of float

Explanation:

8 0

3 years ago

A flywheel having constant angular acceleration requires 4.00 ss to rotate through 164 radrad . Its angular velocity at the end

ExtremeBDS [4]

Answer:

$\omega '=-13.5rad/s$

Explanation:

From the question we are told that:

Time $t=4sec$

Angular displacement $\theta= 161 rad$

Final Angular velocity $\omega = 100 rad / s$

Let

Angular acceleration $\alpha$

Generally the equation for Initial Angular velocity $\omega '$ is mathematically given by

$-\omega '^2=2 \alpha \theta -\omega^2$

$\omega '^2= \alpha 328 +11236$

Also,Initial Angular velocity $\omega '$ is mathematically given by

$\omega '=\omega - \alpha t$

Therefore substitution

$-\omega '^2=2 \alpha \theta -\omega^2$

$\omega '=\omega - \alpha t$

$(\omega - \alpha t)^2=2 \alpha \theta -\omega$

$-16\alpha^2+848\alpha+11236= \alpha 328 -11236$

$16\alpha^2-520\alpha=0$

$\alpha=29.875rads/s^2$

Substitution in the 2nd equation for Initial Angular velocity $\omega '$

$\omega '=106-(29.875rads/s^2*4)$

$\omega '=-13.5rad/s$

6 0

3 years ago

How can I skip class:

Dahasolnce [82]

Answer:

c gl

Explanation:

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3 0

3 years ago

Read 2 more answers

A 4.0-mm-diameter hole is 1.0 m below the surface of a 2.0-m-diameter tank of water. What is the rate, in mm/min, at which the w

kykrilka [37]

Answer:

$1.06$ mm/min

Explanation:

h = height of the water level above the hole = 1 m

$d_{h}$ = diameter of hole = 4 mm = 0.004 m

$A_{h}$ = Area of cross-section of hole = $(0.25) \pi d_{h}^{2}$

$d_{t}$ = diameter of tank = 2 m

$A_{t}$ = Area of cross-section of tank = $(0.25) \pi d_{t}^{2}$

R = rate at which the water level drop

Rate at which the water level drop is given as

$R = \frac{A_{h} (\sqrt{2gh}))}{A_{t}}$

$R = \frac{((0.25) \pi d_{h}^{2}) (\sqrt{2gh}))}{((0.25) \pi d_{t}^{2})}$

$R = \frac{(d_{h}^{2}) (\sqrt{2gh}))}{(d_{t}^{2})}$

$R = \frac{(0.004)^{2} \sqrt{2(9.8)(1)}}{(2)^{2}}$

$R = 17.7\times 10^{-6}$ m/s

$R = 0.0177$ mm/sec

$R = (0.0177) (60)$ mm/min

$R = 1.06$ mm/min

8 0

4 years ago