Physics, asked by nupur90, 1 year ago

Hi ppl....

PHYSICS... CLASS 9
I want ans for 1 , 2 ,3​

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Answers

Answered by Anonymous
6

Explanation:

HI there !!

1)

Displacement is the shortest distance possible between initial and final position.

As the athlete is in circular motion , displacement is zero. [ THis is because , the initial and final position is the same ]

Distance covered in 1 round = Circumference of the circle

Circumference of the circle = 2πr

Diameter = 200 m

Radius = 100 m

Distance covered in 1 round[ 40 sec ] = 2 × 22/7 × 100 = 628.57 m

Distance covered in 1 sec = 628.57 ÷ 40 = 15.71 m

2min 20 sec = 140 sec

Distance covered in 140 sec = 15.71 × 140 = 2199.4 m

2)Velocity = dispacement / time

Speed = distance / time

a) when he jogs from A to B on a straight road,

displacement = distance = 300m

time = 2 minutes 30 seconds = 150 s

velocity = 300/150 = 2 m/s

speed = 300/150 = 2m/s

b)when he jogs from A to B and turns back to C,

displacement = 300-100 = 200m

distance = 300+100 = 400m

time = 3 minute 30 second = 210 s

velocity = 200/210 = 20/21 m/s

speed = 400/210 = 40/21 m/s

3)Let the distance traveled by Abdul from home to school = s km

time taken to reach the school=t1 sec

For Return journey , Abdul cover distance =s km

time=t2 sec

∴ Average speed for forward journey[home - school ] = Total distance/ Total time.

20 km/h =s/t1

∴t1=s/20 h ------eq(1)

Average speed for backward journey[school -home] = Total distance/ Total time.

30 km/h =s/t2

t2=s/30h -----eq(2)

Average distance for entire journey = Total distance/ Total time

=(s+s)/[s/20 +s/30]

=2s/s[1/20+1/30]

=2x20x30/50

=24 km/hr

Shortcut method : If equal distance are covered with speeds v1 and v2 then average distance=2v1v2/v1+v2

= 2x20x30/50

=24km/hr

∴ The average speed for Abdul's trip is 24km/h

hope this answer helpful u

Answered by Anonymous
1

Answer:

❤️❤️❤️ hello ❤️❤️❤️

Explanation:

❤️☺️good evening ☺️ ❤️

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