Explanation:
It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion
Case I
s=ut+
2
1
at
2
10=5u+
2
1
a(5)
2
20=10u+25a
4=2u+5a..............(1)
Case 2
In next 3s the particle covers more 10m distance. So
20=8u+
2
1
a(8)
2
5=2u+8a.........(2)
On solving equation (1) and (2)
4=2u+5a
5=2u+8a
a=
3
1
m/s
2
Put the value of a in equation (1)
u=
6
7
m/s
Now to find distance in next 10 s. total time will be 10s
s=
6
7
×10+
2
1
×
3
1
×(10)
2
s=28.33m
Distance travelled in next 2 sec
s=28.33−20=8.33m
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Given :
Liquid is poured into a burrete so that it reads 14cm³.
50 drops were run each of volume 0.1cm³ .
To Find :
The volume of liquid in burrete after 50 drops.
Solution :
Volume of each drop, v = 0.1 cm³.
Initial volume in burrete, V = 14 cm³.
Now, volume left after droping 50 drops are :

Therefore, the volume left in burrete is 9 cm³ .