A ball is thrown horizontally from a cliff and hits the ground 4 seconds later 40 meters from the base of the cliff. How high wa
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Answer:
= 0.66
Hence, the fraction of the length of the rod above water = = 0.66
and fraction of the length of the rod submerged in water = 1 - = 1 - 0.66 = 0.34
Explanation:
Data given:
Density of the rod = 5/9 of the density of the water.
Let's denote density of Water with w
And density of rod with r
So,
r = 5/9 x w
Required:
Fraction of the length of the rod above water.
Let's denote total length of the rod with L
and length of the rod above with = y
Let's denote the density of rod = r
And density of water = w
So, the required is:
Fraction of the length of the rod above water = y/L
y/L = ?
In order to find this, we first need to find out the all type of forces acting upon the rod.
We know that, a body will come to equilibrium if the net torque acting upon a body is zero.
As, we know
F = ma
Density = m/v
m = Density x volume
Volume = Area x length = X ( L-y)
So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:
F = mg
F = (Density x volume) x g
g = gravitational acceleration
F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)
where,
X (L-y) = volume
w = density of water.
Another force acting upon it is:
F = mg
F2 = X x L x r x g
Now, the torques acting upon the body:
T1 + T2 = 0
F1 ( y + ) g sinФ - F2 x (
) x gsinФ = 0
plug in the equations of F1 and F2 into the above equation and after simplification, we get:
=
. r
where, w is the density of water and r is the density of rod.
As we know that,
r = 5/9 x w
So,
=
. 5/9 x w
Hence,
=
=
Taking common and solving for
, we will get
= 0.66
Hence, the fraction of the length of the rod above water = = 0.66
and fraction of the length of the rod submerged in water = 1 - = 1 - 0.66 = 0.34
Answer:
<em>Explanation below</em>
Explanation:
<u>Speed vs Velocity </u>
These are two similar physical concepts. They only differ in the fact that the velocity is vectorial, i.e. having magnitude and direction, and the speed is scalar, just the magnitude regardless of the direction. They are strongly related to the concepts of displacement and distance, which are the vectorial and scalar versions of the space traveled by a moving object. The velocity can be computed as
Where is the position vector and t is the time. The speed is
To compute , we only need to know the initial and final positions and subtract them. To compute d, we need to add all the distances traveled by the object, regardless of their directions.
Maggie walks to a friend's house, located 1500 meters from her place. The initial position is 0 and the final position is 1500 m. The displacement is
and the velocity is
Now, we know Maggie had to make three different turns of direction to finally get there. This means her distance is more than 1500 m. Let's say she walked 500 m in all the turns, then the distance is
If she took the same time to reach her destiny, she would have to run faster, because her average speed is