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ycow [4]
3 years ago
12

Edward travels 150 kilometers due west and then 200 kilometers in a direction 60° north of west. What is his displacement in the

westerly direction?
Physics
1 answer:
lions [1.4K]3 years ago
8 0
     We must decompose the displacement vector to find the new displacement to the west.

d_{x}=v.cos\O \\ d_{x}=200. \frac{1}{2}  \\ d_{x}=100km
  
     Adding all displacements to West, It comes:

\Delta S= d_{o}+d_{x} \\ \Delta S= 150+100 \\ \boxed {\Delta S= 250 km}

If you notice any mistake in English, please let me know, because I'm not native.

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"a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1
kotegsom [21]

The average angular speed of the grindstone is 10 rad/s

\texttt{ }

<h3>Further explanation</h3>

<em>Let's recall </em><em>Angular Speed</em><em> formula as follows:</em>

\boxed{ \omega = \omega_o + \alpha t }

\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }

\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }

\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }

<em>where :</em>

<em>ω = final angular speed ( rad/s )</em>

<em>ω₀ = initial angular speed ( rad/s )</em>

<em>α = angular acceleration ( rad/s² )</em>

<em>t = elapsed time ( s )</em>

<em>θ = angular displacement ( rad )</em>

\texttt{ }

<u>Given:</u>

radius of the grindstone = R = 4.0 m

initial angular speed = ω₀ = 8.0 rad/s

final angular speed = ω = 12 rad/s

elapsed time = t = 4.0 seconds

<u>Asked:</u>

average angular speed = ?

<u>Solution:</u>

<em>Firstly , we will calculate </em><em>angular displacement </em><em>as follows:</em>

\theta = \frac{( \omega + \omega_o )}{2} t

\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0

\theta = 10 \times 4.0

\boxed {\theta = 40 \texttt{ rad}}

\texttt{ }

<em>Next , we could calculate the </em><em>average angular speed</em><em> as follows:</em>

\texttt{average angular speed} = \theta \div t

\texttt{average angular speed} = 40 \div 4.0

\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Moment of Inertia : brainly.com/question/13796477
  • The Ratio of the Moments of Inertia : brainly.com/question/2176655

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

4 0
3 years ago
Read 2 more answers
Can someone help me due at 1:00 PM plus 100 points <br><br>questions B,C,D
vagabundo [1.1K]
<h3>b. ML </h3><h3>c. M</h3><h3>d. Kg</h3>

<h3>Hope it helps...</h3>
6 0
3 years ago
Read 2 more answers
What is emotional strength?​
Ugo [173]

Answer:

Emotional strength is a person's ability to deal with challenges (any kind) and how they can bounce back on that certain situation.

I hope this helps you.

4 0
3 years ago
The length of the minute hand of the clock is 14cm. Calculate the speed with which the tip of the minute hand moves
Molodets [167]

Speed of the tip of the minute hand=V=0.0244 cm/s

Explanation:

The angular velocity of the minute hand is given by

\omega= \frac{2\pi}{T}

T= time period of the minute hand=60 min=3600 s

so ω= 2 π/3600 rad/s

Now linear velocity v= r ω

r= radius of minute hand=14 cm

so v= 14 (2 π/3600)

V=0.0244 cm/s

8 0
3 years ago
I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an
Nikitich [7]

Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = \sqrt{\frac{T}{\frac{M}{L}}}

Frequency (F₁) = \frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = \frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0
3 years ago
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