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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>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>
Explanation:
The image is shown below.
The force on the particle with charge q due to each charge Q =
we designate this force as N
Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.
Resolving the forces on the particle, we have
for the x-component
= N cosine 60° + (-N cosine 60°) = 0
for the y-component
= -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N
The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( = 0).
The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.
<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>
