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mrs_skeptik [129]
3 years ago
8

As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in

Physics
1 answer:
vlada-n [284]3 years ago
7 0

Answer:

a=-0.33\ m/s^2

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

\displaystyle a=\frac{\Delta v}{t}

Or, equivalently

\displaystyle a=\frac{v_f-v_o}{t}

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2

The negative sing of a indicates there is deceleration or decreasing speed

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Who was the proponent of the Neo-classicism?

a) Claude Debussy

b) Joseph Maurice Ravel

c) Igor Stravinsky

d) Arnold Schoenberg

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2 years ago
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3 years ago
A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hou
Whitepunk [10]

Total distance = 36500 m

The average velocity = 19.73 m/s

<h3>Further explanation</h3>

Given

vo=initial velocity=0(from rest)

a=acceleration= 1 m/s²

t₁ = 20 s

t₂ = 0.5 hr = 1800 s

t₃= 30 s

Required

Total distance

Solution

State 1 : acceleration

\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m

\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s

State 2 : constant speed

\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m

State 3 : deceleration

\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)

\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m

Total distance : state 1+ state 2+state 3

\tt 200 + 36000 + 300=36500~m

the average velocity = total distance : total time

\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s

4 0
2 years ago
How to make a paper airplane curve down and fly upside down?
Ainat [17]
Throw it sideways and try to make it spin around but it needs to be thrown high up then it should kinda glide down
8 0
3 years ago
Read 2 more answers
Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of
IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0
3 years ago
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