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3 years ago

PLEASE HELP

Physics

1 answer:

Elan Coil [88]3 years ago

5 0

Answer:

Explanation:

$a_{x}=0$ $v_{xo}= v_{o}cos(delta)t$ $a_{y}=-g$ $v_{yo}=sin$ θ

X-direction | Y-direction

$x=x_o+v_{xo}t$ ⇒ $x=v_{xo}cos(delta)t$ |

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11. A fundamental property of light is that it: 15

myrzilka [38]

Answer:

Curves around objects

Explanation:

Diffraction is a property of light described by bending of light around an object. This ability of light to bend around edges has facilitated optical effects of light where there is interference of light waves. Other properties of light are: reflection, refraction, polarization, scattering of light, and interference of light.

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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.

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2 years ago

How is a wavelength measured?

kenny6666 [7]

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3 years ago

Read 2 more answers

How is modeling useful in studying the structure of the atom?

Mandarinka [93]

Modelling the structure of the atom is important because modeling replaces the real system with something similar but easier to examine. Option B

<h3>What is modeling?</h3>

A model is a representation of reality. We know that a model could help us to recreate reality in a manner that we could be able to relate fully with it. A model could be used also a means of explanation.

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1 year ago

An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and

Svetradugi [14.3K]

Answer:

Total contraction on the Bar = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm

Diameter for aluminum bar = 40 mm

Hole diameter = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180 × 10³ N

Calculate the total contraction on the bar = ???

The relation used in calculating the contraction on the bar is:

$\delta L = \dfrac{P *L }{A*E}$

The relation used in calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Total contraction on the Bar = $\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]$

Total contraction on the Bar = 2.117( 0.398 + 0.182)

Total contraction on the Bar = 2.117*(0.58)

Total contraction on the Bar = 1.22786 mm

5 0

3 years ago