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2 years ago

Describe the motion represented by a horizontal line on a distance time graph

2 answers:

8 0

A horizontal line on a distance time graph is saying that the distance doesn't change as time goes on. This tells us that the object's speed is zero, and its velocity is zero, which pretty much says that the object is stationary ("at rest").

MissTica2 years ago

6 0

A horizontal line on a distance-time graph indicates that an object is stationary.

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A 95 kg fullback, running at 8.2 m/s, collides in midair with a 128 kg defensive tackle moving in the opposite direction. Both p

Sindrei [870]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below is the answers:

Fullback running

Mo = mass * velocity 
Mo = 95kg * 8.2 m/s =779 kg*m/s (a 

He got stopped Change in Mo = 779 kg*m/s (b 

Both stopped ===> Tackle's mo = - Halfback's Mo = - 779 kg*m/s (c & d 

- 779 = 128 * v 
v= - 6.09 m/s (e

6 0

2 years ago

What voltage would produce a current of 100amps through an aluminum wire assuming that the re sistance of the wire is3.44×10-4 o

agasfer [191]

V = IR

I = current
R = resistance

Voltage = 100 * (3.44x 10^-4) = do the calculation

Hope this helps

8 0

2 years ago

How many grams are in 6.53 moles of Mn?

Westkost [7]

Answer:

359 g Mn

General Formulas and Concepts:

Dimensional Analysis
Reading the Periodic Table of Elements

Explanation:

Step 1: Define

6.53 mol Mn

Step 2: Find conversion

1 mol Mn = 54.94 g Mn

Step 3: Dimensional Analysis

$6.53 \hspace{3} mol \hspace{3} Mn(\frac{54.94 \hspace{3} g \hspace{3} Mn}{1 \hspace{3} mol \hspace{3} Mn} )$ = 358.758 g Mn

Step 4: Simplify

We are given 3 sig figs.

358.758 g Mn ≈ 359 g Mn

3 0

2 years ago

Initially, a particle is moving at 5.25 m/s at an angle of 35.5Â° above the horizontal. Three seconds later, its velocity is 6.0

ivolga24 [154]

Answer:

a =( -0.32 i ^ - 2,697 j ^) m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

v₀ₓ = v₀ cos θ

v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

$v_{oy}$ = v₀ sin θ

$v_{oy}$ = 5.25 sin35.5

$v_{oy}$ = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

$v_{y}$ = v₀ sin θ

$v_{y}$ = 6.03 sin (-56.7)

$v_{y}$ = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

a = ( $v_{f}$ - v₀) /t

aₓ = (3.31 -4.27)/3

aₓ = -0.32 m/s²

$a_{y}$ = (-5.04-3.05)/3

$a_{y}$ = -2.697 m/s²

6 0

3 years ago

Any 3 differences between telescope and microscope

Keith_Richards [23]

An instrument used to observe or imagine very small object using an optical mangifier
mirco cell.
Telescope is a magnifer of distance object

4 0

3 years ago