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raketka [301]
2 years ago
10

--->Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t

he
direction along which Q is flying. Find the magnitude of the
velocity of the aircraft P relative to aircraft Q.

pls guys I need answer so answer to me if you know the answer pls pls pls​
Physics
1 answer:
REY [17]2 years ago
5 0

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

                                  <em> V = v cos Ф </em>

Substituting the values in the above equation

                                   v = 300 x cos 90°

                                      = 300 x 0

                                      = 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

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A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s
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Hence, the radiation pressure is:

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A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal
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(a) 18.9 m/s

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- A vertical motion with constant acceleration (g=9.8 m/s^2) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

v=18.9 m/s

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

v_y^2 - u_y^2 = 2ah (1)

where

u_y = 16.4 m/s is the initial vertical velocity

v_y is the vertical velocity at height h

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, v_y = 0, so we can use the equation to find the maximum height

h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m

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v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s

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v_2 = 22.2 m/s

So the difference is

\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s

And so, in percentage,

\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%

So, the stone in part (b) is moving 17.4% faster than in part (a).

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