Which one of the following acts is an example of body language?

Physics

2 answers:

Ludmilka [50]3 years ago

5 0

Answer:

C. rolling of the eyes

Explanation:

(◔_◔)

just olya [345]3 years ago

4 0

An example of body language would be rolling your eyes. That's saying that you're bored, don't care, stop talking to other people without even opening your mouth.

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An airplane is heading south at a speed of 600km/h. If a wind begins blowing from the southwest at a speed of 100km/h, calculate

Karo-lina-s [1.5K]

Answer:

Here's what I got:

Let's assume that N and E are + directions while S and W are - directions.

Wind is blowing from SW; thus, it is blowing towards NE (or at 45 deg N of E).

Dividing the wind's speed into components:y-component: +70.71 km/h; x-component: +70.71 km/h

Dividing the airplane's speed into components:y-component: -600 km/h; x-component: 0 km/h

Adding the components to get the resulting components:y-component: -529.29 km/h; x-component: +70.71

Using the Pythagorean Theorem to find the resulting speed:v^2 = y^2 + x^2 so v = 533.99 km/h

To find the angle of direction, use arctan (y/x):arctan (529.29/70.71) = 82.39 deg

ANSWER: velocity = 533.99 km/h at 82.39 deg S of E

Explanation:

8 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).