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Answers
Answer:
This was quite much! IT TOOK ME SO MUCH TIME!!
I didn't do solve this I found these answers expert verified when I searched for the questions.
I really hope this helps so much!!
Explanation:
1.Given:
Initial velocity= 36km/h=36x5/18=10m/s
Final velocity =54km/h=54x5/18=15m/s
Time =10sec
Acceleration = v-u/ t
=15-10/10=5/10=1/2=0.5 m/s2
Distance =s
From second equation of motion:
S=ut +1/2×at² (ut= initial velocity at²= acceleration × time²)
=10*10+1/2*0.5*10*10
=100+25
=125m
So distance traveled is 125m
2.
initial velocity = 10m/s (change from km/hr to m/s)
final velocity = 15m/s
DECELERATION = (initial - final velocity) / time
time given = 10 sec
deceleration = 10 m/s - 15 m/s / 10 sec
deceleration = -5/10
RETARDATION = -0.5
3.
1) displacement = 0
distance=22/7x20m
2) displacement= 20m
distance=(1.5)22/7x20
3) displacement=0
distance=(2) 22/7x20m
4) displacement=20m
distance=(2.5) 22/7x20m
4.
Given; u=0
a=5m/s
t=5sec.
v=u+at
V=0+5*5
V=25 m/s.
5.Given:
Maximum speed= 90km/hr
Time=10hrs
DISTANCE = 500km
Average speed= TOTAL DISTANCE/ TOTAL time
= 500/ 10
=50km/ hr
Ratio of maximum speed and average speed is = 90/ 50
=9:5
Ratio of maximum speed and average speed is =9:5
6. As the car is starting from rest ,
Initial velocity = u = 0 m/s
Final velocity = v = 54 km/hr
To convert km/hr to m/s , we multiply it by 5/18
Final velocity = v = 54 km/hr = 54 x 5/18 = 15 m/s
Time (t) = 2 sec
a = acceleration
Equation of motion :
v = u + at
15 = 0 + 2a (We r finding aceleration(a))
a = 15/2 = 7.5 m/s²
(i) Acceleration = 7.5 m/s²
Equation of motion :
s = distance traveled
s = ut + 1/2 at²
s = 0 + 1/2 × 7.5 × 2²
(ii) Distance Travelled = 15 metres
7.1) What is the net displacement?
Ans. Displacement for the above situation is 0. As we know, that displacement is the shortest path from the initial to the final point. Here, the initial and the final points are the same, and hence, it takes no time to travel. So, the displacement is 0.
2) What is the distance?
Ans. Distance for the above situation is 200 m. Here, the ball has to cover a distance of 100 m and while coming back, again it has to cover the same distance, hence, the distance to be covered here is 200 m.
Distance =100 + 100 = 200