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Kamila [148]
3 years ago
7

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 \rm {km/hour} is blowing sout

hward. The speed of the plane relative to the air (called the "air speed") as measured by instruments aboard the plane is 310 \rm {km/hour}. In which direction should the pilot head?
Part A: What is the speed of the air relative to a person standing on the ground, v_A/G? What is the speed of the plane relative to the air, v_P/A?
Physics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

θ = 14°    ........ North of West

V_A/G = 300 km/h   ........ West

Explanation:

Given:

- The speed of wind V_A = 75 km/h ( South )

- The speed of plane relative to air V_P/A = 310 km/h

- The plane wants to go in westwards.

Find:

In which direction should the pilot head?

What is the speed of the air relative to a person standing on the ground, v_A/G?

Solution:

- The plane wants to go westwards; however, the wind would push the plane down south. To combat the effect of wind the plane needs to travel somewhat North West just enough such that wind pushes it down to a point westwards. The angle the plane travels North of west is θ.

- Construct a velocity vector triangle on a coordinate system where unit vectors +i = East , -i = West , +j = North and -j = South. With plane's initial position as origin.

- We know that the plane travels relative to air at angle θ North of west we have the following velocity vector for V_P/A:

                      vector (V_P/A) = -V_P/A*cos(θ) i + V_P/A*sin(θ) j

- Similarly for velocity of wind V_w and plane relative to ground or stationary observer on ground V_A/G is:

                      vector (V_w) = -V_w j = -75 j km/h

                      vector (V_A/G) = -V_A/G i = -V_A/G i km/h

- Use the relative velocity formulation:

                      vector (V_A/G) = vector (V_P/A) + vector (V_w)

- Plug the respective expressions developed above:

                     -V_A/G i km/h = -310*cos(θ) i + 310*sin(θ) j -75 j km/h

- Now compare coefficients of i and j unit vectors we have:

 For j unit vector:  310*sin(θ) -75  km/h = 0  

                        sin(θ) = 75 / 310 = 0.2419354839

                        θ = arcsin(0.2419354839)

                        θ = 14°    ........ North of West

 For i unit vector:  -V_A/G = -310*cos(θ)  

                        V_A/G = 310*cos(14)                                

                        V_A/G = 300 km/h   ........ West

Answer: The plane must travel at 14 degrees north of west if it wants to end up at any point west of its direction. The stationary observer at ground will see the plane moving west at a speed of 300 km/h.

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Answer:

0.05 W

Explanation:

The power dissipated by a device can be written as

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What is the kinetic energy in joules of a 870-kg automobile traveling at 95 km/h?
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Answer:

The kinetic energy is 303178 Joule

Explanation:

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1h ----- 3600 seconds

v = (95km 1000m / km) / (1h x 3600s / h) = 26, 4 m / s

We use the Kinetic Energy formula: Ec = 1/2 m x v ^ 2

Ec = 1/2 x 870 kg x (26, 4) ^ 2 = 303177, 6 Joule

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A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = (1⁄4u2) N, where u is in radians, is applied to t
Soloha48 [4]

A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = (1⁄4θ²) N, where θ is in radians, is applied to the cord, determine the disk’s angular acceleration when it has turned 5 revolutions. The disc has an initial angular velocity \omega _o = 1 \ rad/s and radius from the center of the disc = 300 mm

Answer:

the angular acceleration = 205.706 rad/sec²

Explanation:

GIVEN THAT:

The disc mass = 8 kg

Force = \dfrac{1}{4} \ \ \theta ^2* N

We are told that the given θ is in radians; Therefore; the when it has turned 5 revolutions; we have the θ to be:

\theta = 5 rev * (\dfrac{2 \  \pi  * rad}{1 rev}) \\ \\ \theta = 10 \ \pi \ rad

Also;

the initial angular velocity \omega _o = 1 \ rad/s

radius from the center of the disc = 300 mm = 0.3 m

Thus; the mass moment about the Inertia can be determined via the following expression;

I_o = \dfrac{1}{2}*m*r^2

I_o = \dfrac{1}{2}*8*0.3^2

I_o = 0.36 \ kg/m^3

Now to calculate the angular acceleration; we equate the sum of the moments acting on the Inertia;

SO:

\sum M_o = I_o \alpha

F*0.3 = 0.36* \alpha

\dfrac{1}{2}* \theta^2 *0.3 = 0.36* \alpha

\alpha = 0.20836 \  \theta^2 \ rad/sec^2

\alpha = 0.20836 \  (10 \ \pi )^2 \ rad/sec^2

\alpha = 205.706 \ rad/sec^2

Hence; the angular acceleration = 205.706 rad/sec²

6 0
3 years ago
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