PLEASE HELP!!!!

Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude

natka813 [3]

Answer:

For vector u, x component = 10.558 and y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So $u_x=ucos\Theta =16.6\times cos50.5=10.558$

Vertical component $u_y=usin\Theta =16.6\times sin50.5=12.808$

So vector u will be u = 10.558 i+12.808 j

Unit vector $u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j$

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component $v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316$

Vertical component $v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464$

So vector v will be 23.6316 i - 6.464 j

Unit vector of v $=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j$

8 0

2 years ago

PLEASE HELP! I NEED TO ANSWER THIS QUICKLY!

rodikova [14]

Answer:

I'm not a genius ok?

Explanation:

1. Radar communication, Analysis of the molecular and atomic structure, telephone communication

2. c

4 0

2 years ago

What were the two classifications of motion according to Aristotle?

wel

Answer:

natural motion and violent motion

Explanation:

8 0

2 years ago

You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the

lana [24]

Answer:

<em>The fifth option is the correct answer: mv; 2 mv</em>

Explanation:

<u>Change of Momentum</u>

Assume an object has a momentum p1 and after some interaction it now has a momentum p2, the change of momentum is

$\Delta p=p_2-p_1$

The momentum is computed as

$p=mv$

Where m is the mass of the object and v its speed. Now let's analyze the situation of both the ball and the clay.

The clay has an initial speed v and a mass m, thus its initial momentum is

$p_1=mv$

When it hits the wall, it sticks, thus its final speed is 0 and

$p_2=0$

The change of momentum is

$\Delta p=0-mv=-mv$

The absolute change is mv

Now for the ball, the initial condition is the same as it was for the clay, but the ball hits back at the same speed, thus its final momentum is

$p_2=-mv$

The change of momentum is

$\Delta p=-mv-mv=-2mv$

The absolute change is 2mv

The fifth option is the correct answer: mv; 2 mv

3 0

3 years ago

A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?

Wewaii [24]

<h2>Hello!</h2>

The answer is: 19.59 m

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

$Vf^2=Vi^2+2*g*s$

Where:

Vf= Final velocity = 0

Vi= Initial velocity = $\frac{19.6m}{s}$

g = Gravity Acceleration = $\frac{9.81m}{s^{2} }$

s = Traveled distance

$0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m$

Have a nice day!

8 0

2 years ago