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gladu [14]
4 years ago
12

Two racecars are driving at constant speeds around a circular track. both cars are the same distance away from the center of the

track, but car 2 is driving twice as fast as car 1. the acceleration of car 2 is ___________ the acceleration of car 1.
Physics
1 answer:
emmainna [20.7K]4 years ago
8 0
If the track has circumference c miles, then and the car's speeds are x and y mi/s, then since time = distance/speed, 

<span>In the same direction, the slow car has a "lead" of 1 mile which the faster car has to make up. </span>

<span>1/(y-x) = 120 </span>
<span>1/(x+y) = 30 </span>

<span>120(y-x) = 30(x+y) </span>
<span>120y-120x = 30x+30y </span>
<span>90y = 150x </span>
<span>y = 5/3 x </span>


<span>1/(x + 5/3 x) = 30 </span>
<span>8/3 x * 30 = 1 </span>
<span>80x = 1 </span>
<span>x = 1/80 mi/sec = 45 mph </span>
<span>y = 75 mph</span>
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3 years ago
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Answer:

\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

<u>Finding </u><u>the</u><u> </u><u>velocity</u><u> </u>

\boxed{ \sf{velocity =  \frac{distance \: travelled \: }{time \: taken}}}

\dashrightarrow{ \sf{velocity =  \frac{200 \: m}{24 \: s} }}

\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}

Hope I helped!

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3 0
3 years ago
7. CALCULATE: How much work is done in each of
kifflom [539]
<h3>Answer</h3>

a. 8J

b. 90J

<h3>Notes formula</h3>

W = F × s

W = Work done (J)

F = Force (N)

s = distance (m)

<h3>Known that </h3>

a. F = 4N

s = 2m

b. F = 30N

s = 3m

<h3>Question</h3>

a. W = ..?

b. W = ..?

<h3>Way to do it</h3>

a. W = F × s

= 4N × 2m = 8J

b. W = F × s

= 30N × 3m = 90J

<em>#Moderators please don't be mean, dont delete my answers just to get approval from your senior or just to get the biggest moderation daily rank.</em>

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