How do I find the direction of the rotation and the magnitude

Can somebody help me out with this please?

Yuki888 [10]

7 I hope this helps!

7 0

4 years ago

A carnival charges S5 for admission plus $2 per ride. Write an expression for the cost of admission plus r rides. How many rides

Airida [17]

Answer : 5 rides

Explanation

15=5+2r
-5 -5
10=2r
/2 /2
5=r

8 0

4 years ago

A car travels 32 km due north and then 46 km in a direction 40° west of north. Find the direction of the car's resultant vector.

tatuchka [14]

Answer:

Step-by-step explanation:

This requires some serious work before we even begin. First off, we will convert the km to meters:

32 km = .032 m

46 km = .046 m

And then we have to deal with the angle given as 40 degrees west of north. An angle 40 degrees west of north "starts" at the north end of the compass and moves towards the west (towards the left in a counterclockwise manner) 40 degrees. That means that the angle that is made with the negative x axis is a 50 degree angle. BUT the way angles are measured in standard form are from the positive x-axis, therefore:

40 degrees west of north = 50 degrees with the negative x axis = 130 degrees with the positive x axis. 130 is the angle measure we use. Phew! Now we're ready to start. Adding vectors requires us to use the x and y components of vectors in order to add them.

$A_x=.032cos90.0$ so

$A_x=0$ (the 90 degrees comes from "due north")

$B_x=.046cos130$ so

$B_x=-.030$ and if we add those to get the x component of the resultant vector, C:

$C_x=-.030$ And onto the y components:

$A_y=.032sin90.0$ so

$A_y=.032$

$B_y=.046sin130$ so

$B_y=.035$ and if we add those together to get the y component of the resultant vector, C:

$C_y=.067$ Note that since $C_x$ is negative and $C_y$ is positive, the resultant angle (the direction) will put us into QII.

We find the magnitude of C:

$C_{mag}=\sqrt{(-.030)^2+(.067)^2}$

We will round this after we take the square root to the thousandths place.

$C_{mag}=.073m$ and now for the angle:

$\theta=tan^{-1}(\frac{.067}{-.030})$ which gives us an angle measure of -67, but since we are in QII, we add 180 to that to get that, in sum:

The magnitude of the resultant vector is .073 m at 113°

6 0

3 years ago

Explain in detail how to graph y=-3/4 x+5

Leviafan [203]

You will start the 5 on the Y axis
from there go down 3 and to the right 4 times

4 0

3 years ago

A projectile is fired at such an angle that the vertical

svetoff [14.1K]

(a) The projectile remains in the air for 10 seconds

(b) It travels a horizontal distance of 600 meters

Step-by-step explanation:

A projectile is fired at such an angle that

1. The vertical component of its velocity is 49 m/sec

2. The horizontal component of its velocity is 60 m/sec

We need to find:

(a) How long the projectile remains in the air

(b) The horizontal distance it travels

∵ The vertical distance y = $u_{y}$ t - $\frac{1}{2}$ g t², where

$u_{y}$ is the vertical component of its velocity, g is the acceleration

of gravity and t is the time

∵ y = 0 ⇒ it return to the same initial height

∵ $u_{y}$ = 49 m/s

∵ g = 9.8 m/s²

- Substitute these values in the rule above

∴ 0 = 49 t - $\frac{1}{2}$ (9.8) t²

∴ 0 = 49 t - 4.9 t²

- Take t as a common factor

∴ 0 = t (49 - 4.9 t)

- Equate each term by 0

∴ t = 0 ⇒ at initial position

∴ 49 - 4.9 t = 0

- Add 4.9 t to both side

∴ 49 = 4.9 t

- Divide both sides by 4.9

∴ t = 10 seconds

(a) The projectile remains in the air for 10 seconds

∵ The horizontal distance = $u_{x}$ t, where $u_{x}$ is the

horizontal component of its velocity

∵ $u_{x}$ = 60 m/sec

∵ t = 10 seconds

∴ x = 60 × 10 = 600 meters

(b) It travels a horizontal distance of 600 meters

Learn more:

You can learn more about the component of velocity in brainly.com/question/4464845

#LearnwithBrainly

3 0

3 years ago

How do I find the direction of the rotation and the magnitude ​

How do I find the direction of the rotation and the magnitude