Given :
Walk in forward direction is 30 m .
Walk in backward direction is 25 m .
To Find :
The distance and displacement .
Solution :
We know , distance is total distance covered and displacement is distance between final and initial position .
So , distance travelled is :
D = 30 + 25 m = 55 m .
Now , we first move 30 m in forward direction and then 25 m in backward direction .
So , displacement is :
D = 30 - 25 m = 5 m .
Therefore , distance and displacement covered is 55 m and 5 m respectively .
Hence , this is the required solution .
<h2>Answer:</h2><h2>The depth of barge float=
3 cm</h2><h2>
Explanation:</h2>
Length of rectangular barge=5.2 m
Width of rectangular barge=2.4m
Mass of crate=410 kg
Let h be the height of barge float
Volume of barge float=
Density of water=
Weight of water displaced by barge=Buoyant force=-Weight of horse



1 m=100 cm
cm
Hence, the depth of barge float=3 cm
<h2 />
Part a)
in horizontal direction there is no gravity or no other acceleration
so in horizontal direction the speed of clam will remain same

Part b)
In vertical direction we can use kinematics



part c)
if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed
Answer:
The answer is 218
Explanation:
Weight = mass * gravitational acceleration
weight is represented by F
F = 25kg (8.7)
(I'm pretty sure that you don't have to include the meters per second/per second thing)
Answer:
Explanation:
From A to B
distance traveled with velocity
in time
from B to C
distance traveled is 0.5 d with
and
velocity for half-half time
divide 1 and 2 we get
Now average velocity is given by
taking
common