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3 years ago

Just need help with 1 and 2 please :D i’m having a bit of trouble :/

Physics

1 answer:

dexar [7]3 years ago

7 0

1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.

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Answer:

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2 years ago

B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

$V_{int}=\sqrt{U^2+19.6H}$

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration is mathematically given as

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Where

acceleration is still -9.81 m/s2,

Hence,

$V^2-U^2=2(-9.81)*-H$

Therefore

$V_{int}=\sqrt{U^2+19.6H}$

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2 years ago

A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (

Margarita [4]

$\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}$

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$\large\displaystyle\text{$\begin{gathered}\sf 1/4 \ mile = 402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \ V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \ in \ m/s} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}$

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2 years ago

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strojnjashka [21]

Answer:

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