Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?

A Moderate Wind Accelerates A Pebble Over A Horizontal Xy Plane With A Constant Acceleration A With Arrow 460 Ms2i Hat 700 Ms2j At Time T 0 The Velocity Is 43 M class=

Sagot :

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]

Velocity acquired during this time

[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]

Consider vertical motion

[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]

Net velocity is

[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]

Angle made is

[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.