Ask question

Ask question

All categories

2 years ago

6

a rural mail carrier leaves the post office and drives 22.0 km in a Northely direction. then drives in a 60° south of east direc

tion for 47 km. what is the displacement from the post office?

1 answer:

Scrat [10]2 years ago

7 0

Answer:

<em>Good Luck!</em>

<em> 5 Stars</em>

<em />

Explanation:

You might be interested in

You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w

Vesna [10]

Answer:

$\ \text{m/s}$

Explanation:

$u_1$ = Velocity of one lump = $3x+3y-3z$

$u_2$ = Velocity of the other lump = $-4x+0y-4z$

m = Mass of each lump = $30\ \text{g}$

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

$mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}$

The velocity of the stuck-together lump just after the collision is $\ \text{m/s}$ .

4 0

3 years ago

cup of hot chocolate has a spoon in it. How does the heat move from the hot chocolate to the metal spoon handle?

serg [7]

The heat moves from the hot chocolate to the handle of the spoon by a process called thermal conduction.

It is the transfer of heat energy from one object to another when they are in contact with eachother.

Hope this answers your question.

5 0

3 years ago

A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

1 year ago

Find the y-component of this

Alborosie

Answer:

-0.0789 m

Explanation:

Recall that the y-component comes associated with the sin(18.4) through the following trigonometric relationship:

y = 0.250 sin(-18.4) ≈ -0.0789 m

Notice it is negative since it is below the x-axis.

4 0

3 years ago

The unusually bright centers found in some galaxies are called ______.

Nimfa-mama [501]

<span>Active Galactic Nuclei.</span>

4 0

3 years ago