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MakcuM [25]
3 years ago
6

A jet airliner moving initially at 548 mph

Physics
1 answer:
sasho [114]3 years ago
3 0

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

v_1 =548 mph along the x-direction

v_2 = 343 mph in a direction 67^{\circ} north of east.

To find the resultant, we must resolve both vectors on the x- and y- axis:

v_{1x}= 548 mph

v_{1y}=0

v_{2x} = (343 mph)( cos 67^{\circ})=134.0 mph

v_{2y} = (343 mph)( sin 67^{\circ})=315.7 mph

So, the components of the resultant velocity in the two directions are

v_{x}=548 mph+134 mph=682 mph

v_{y}=0 mph+315.7 mph=315.7 mph

So the new speed of the aircraft is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(682 mph)^2+(315.7 mph)^2}=751.5 mph

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Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

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Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹  N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹  N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea =  +1.8 * 10⁴ N/C

Electric field due to charge B

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Eb =  -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

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E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

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