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Allushta [10]
3 years ago
11

An airplane flies at 100km/h in still air. If it flies into a 10 km/h headwind, its groundspeed is

Physics
2 answers:
HACTEHA [7]3 years ago
5 0

Answer:

D) 90\frac{km}{h}

Explanation:

The groundspeed of the airplane is equal to the speed of the airplane in still air, plus the speed of the air. Since the plane flies with headwind, the sign of the air speed is negative, since it is contrary to the movement of the airplane.

Now, we have:

V_g=V_a+V_h

Here V_g is the airplane groundspeed, V_a the airplane speed and V_h the air speed

V_g=100\frac{km}{h}+(-10\frac{km}{h})=90\frac{km}{h}

timama [110]3 years ago
3 0
Its D.) 100-10 is 90 the plane is flying  in the 10 km/h wind at 100 km/h take away the resistance and you get your ground speed 

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A doctor has a patient blow on one side of a U-shaped tube that is partially filled with water and has a ball floating on the ot
natali 33 [55]

The patient apply pressure to his side of the tube is 15 x 10⁴ N/m². The correct option is D.

<h3>What is pressure?</h3>

The pressure is the amount of force applied per unit area. It is represented as

Pressure p = Force/Area

Given is a doctor has a patient blow on one side of a U-shaped tube that is partially filled with water and has a ball floating on the other side. The ball rises with a force of 150 N. The cross section are of the tube is 0.001 m³

The pressure applied is

p = 150 /0.001

p = 15 x 10⁴ N/m²

Hence, the patient apply pressure to his side of the tube is 15 x 10⁴ N/m².

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6 0
2 years ago
A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a
vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451

tan\alpha = 0.451

\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0

3 0
4 years ago
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