Can someone give me the units

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

A man can jog 10 miles in 90 minutes. What’s his speed in mph?

Yuki888 [10]

Answer:

hi there!

the correct answer to this question is: 6.67 mph

Explanation:

you convert minutes to hours

10 miles * 60 mins / 90 mins

7 0

2 years ago

The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in

MissTica

Answer:

Explanation:

Using the equation of motion to find the angular acceleration:

$\omega_f = \omega_i + \alpha t$

$\omega_f$ is the final angular velocity in rad/s

$\omega_i$ is the initial angular velocity in rad/s

$\alpha$ is the angular acceleration

t is the time taken

Given the following

$\omega_f = 6100rpm$

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x = 638.792rad/s

$\omega_f = 638.792rad/s\\$

Get the angular acceleration:

Recall that:

$\omega_f = \omega_i + \alpha t$

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

5 0

2 years ago

Which of the following can be the first step to take in organizing data

Natalka [10]

the first step would be gathering information , and making sure its correct

5 0

3 years ago

An ac power generator produces 73.37 a (rms) at 4623 v. the voltage is stepped up to 105,033 v by an ideal transformer, and the

evablogger [386]

Step up transformer is a device which is used to step up the voltage which is input with some value.

This is based upon the principle of mutual inductance and in this the voltage input and voltage output is different because of number of turns.

Here if ideal transformer is given then power input and power output of the transformer must be same as there is no power loss in ideal transformer.

So we can write

$i_pV_p = i_sV_s$

here

$i_p$ = 73.37 A

$V_p$ = 4623 V

$V_s$ = 105033 A

now using above equation we will have

$73.37*4623 = 105033*i_s$

solving above we will have

$i_s = 3.23 A$

7 0

2 years ago