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
Answer:
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The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.
The distance of car from the intersection point after t hours is 
.
The distance of truck from the intersection point after t hours is 
.
Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is
Thus the distance apart is 
there’s no photo. brainly hasent been working for me today so not sure
Answer:
the pressure due to the water on the diver is 200,000 pascal
pressure = height × density × acceleration due to gravity
p = 20×1000×10
p=200,000 pascal
