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3 years ago

How do birds manage to fly without flapping their wings?

1 answer:

3 0

It's called gliding.
So, they will get up high and collect their balance and be nice and steady. Then, they can spread their wings out and glide. The wind will carry them from their.
How do we manage to stand without constantly walking? It's the same exact thing.

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. A bird is flying with a speed of 18.0 m/s over water when it accidentally drops a 2.00 kg fish. If the altitude of the bird is

zysi [14]

V^2 = u^2 + 2gr, where v - speed, u - initial speed=0, r - displacement (or height)
v^2 = 0 + 2*10*5.4
v = 10.2 m/s

4 0

3 years ago

Read 2 more answers

Two long, parallel wires are separated by a distance of 2.00 cm . The force per unit length that each wire exerts on the other i

RUDIKE [14]

Answer:

a) current in the second wire is 5.60A

b) opposite directions

Explanation:

a) We need to find the current of wire, the magnitude of the force per unit length between the two wires carrying current I and I¹ is given by

$\frac{F}{L} = \frac{U_0I^1}{2\pi r} \\\\I^1 = \frac{F2\pi r}{LU_0I}$

$= (3.6 * 10^-^5)\frac{2\pi (0.02)}{4\pi * 10^-^7)(0.64)} \\= 5.60A$

b) knowing that for a two parallel conductor carrying current in the same direction attracts each other, and for a two parallel conductors carrying carying current in opposite direction repels eachother.

therefore, since the two wire repel each other then the current in the second wire must flow in the opposite direction of the current in the first wire.

7 0

4 years ago

Please helo me to 1st question

k0ka [10]

Explanation:

6 min in seconds:

6×60

<h2>360 seconds </h2><h2 /><h2 />

6 min in hours:

<h2>0.1 hours</h2>

6 0

3 years ago

How many moons does jupiter have?

lidiya [134]

Jupiter has 49 official, named moons and 14 more unofficial ones.

4 0

3 years ago

Read 2 more answers

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago