A change in position with respect to a reference point is called?

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago

Which of the following is an example of how humans can increase biodiversity?

Leto [7]

Answer:

The following is an example of how humans can increase biodiversity

Provide Wildlife Corridors and Connections Between Green Spaces

Use Organic Maintenance Methods and Cut Back On Lawns

Use a Native Plant Palette and Plant Appropriately

Utilize Existing Green Space Connections

Be Mindful of Non-Native Predators

7 0

3 years ago

7. Katie and her best friend liam play tennis every saturday morning. When katie serves the ball to liam, it travels 9.5 meters

Ksenya-84 [330]

Answer:

The velocity of the tennis ball is 4.52 m/s.

Explanation:

Given that,

The distance covered by ball, d = 9.5 meters (due south)

Time, t = 2.1 sec

Let v is the velocity of the tennis ball. We know that the velocity of an object is given by the total distance covered divided by total time taken. It is given by :

$v=\dfrac{d}{t}\\\\v=\dfrac{9.5\ m}{2.1\ s}\\\\v=4.52\ m/s$

So, the velocity of the tennis ball is 4.52 m/s. Hence, this is the required solution.

6 0

3 years ago

1) Si un mango cae a una velocidad de 75m/s y tarda 26 seg. en caer. ¿ Cuál habrá sido la velocidad con qué el mango llegó al su

Lyrx [107]

Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Explanation:

El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:

$v = v_{o}+g\cdot t$ (1)

Donde:

$v_{o}$ - Velocidad inicial, en metros por segundo.

$v$ - Velocidad final, en metros por segundo.

$g$ - Aceleración gravitacional, en metros por segundo al cuadrado.

$t$ - Tiempo, en segundos.

Si sabemos que $v_{o} = -75\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ y $t = 26\,s$ , entonces la velocidad final del mango es:

$v = v_{o}+g\cdot t$

$v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)$

$v = -329.982\,\frac{m}{s}$

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

8 0

2 years ago

Your roommate is working on his bicycle and has the bike upside down. He spins the 60 cm

tigry1 [53]

Diameter = 60 cm, Radius = 60/2 = 30 cm = 30/100 = 0.3 m.

The pebble in the tread goes by 3 times every second.

This is the same as 3 times per second.

Recall the unit of frequency is Hertz or per second, s⁻¹

So 3 times per second, Frequency, f = 3s⁻¹ or 3 Hertz

For angular motion:

Angular speed, ω = 2πf

   = 2*π*3

   = 6π   rad/s

Linear speed, v = ωr = 6π * 0.3 = 1.8π m/s

Linear acceleration, a = v² / r

   a = 1.8π * 1.8π / 0.3 = 10.8π² m/s²

Angular acceleration α = a/r = 10.8π² / 0.3 = 36π² rad/s²

Angular speed = 6π rad/s ≈ 18.840 rad/s

The linear speed of the pebble = 1.8π m/s ≈ 5.655 m/s

The angular acceleration = 36π² rad/s² ≈ 355.306 rad/s²

The linear acceleration of the pebble = 10.8π² m/s ≈ 106.592 m/s²

5 0

3 years ago