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Let two uniform rods each of mass m and length l joined at angle #theta# constitute the V shaped rigid body ABC. The rigid body is suspended freely from point A and its BC part becomes horizontal and AB part remains inclined at angle #theta# with the horizontal when the system comes in equilibrium.

Let the midpoint of AB be P and that of BC be Q.The vertical line drawn through the point of suspension A intersects BC rod at O.The weight of AB part #mg# will act vertically downward through P and its line of action intersects BC at R. Again weight (mg) of BC will act downward through its mid point Q .These two weights will lie in two opposite sides of O

The perpendicular distannce of the weight acting through P from O is #OR=l/2costheta#.
Again the perpendicular distance of the weight acting through Q from O is

#OQ=BQ-BO=l/2-lcostheta#

So at equlibrium of two weigts

#mgxxOR=mgxxOQ#

#mgxxl/2costheta=mgxx(l/2-lcostheta)#

#=>3/2costheta=1/2#

#=>costheta=1/3#

#=>theta=cos^-1(1/3)= 70.5^@#

Answer:

#W=1.7J#

Explanation:

To calculate the work required to extend the spring 93 cm from its equilibrium position given the spring constant, we can consider the area under a force (#F_(sp)#) vs displacement (#x#) graph. The slope of this graph is #k#, the spring constant, and the area under the "curve" is work. This is given by Hooke's Law, where #F_(sp)=kx#. Because this is a linear relationship (constant force along a straight line), the "curve" is actually a line.

BBC.co.uk

You can find the area under the force vs. displacement graph with basic calculus, or perhaps you were taught to use the relationship #W=1/2(k)x^2#. Either will give you the same answer, and I'll demonstrate both.

Using #W=1/2kx^2#:

#W=1/2(4(kg)/s^2)(.93m)^2#

#W=1.7J#

Using calculus:

#W=int_(s_i)^(s_f)kxdx#

Where #s_i# is the spring's initial displacement from equilibrium and #s_f# is its final displacement, i.e. how far it was stretched. Because we have assumed that the spring is stretched from equilibrium, its initial displacement from its equilibrium position is 0, and then it was stretched .93 m from that position.

#W=int_(0)^(.93)4xdx#

#W=2x^2#

Evaluating for the limits of integration (#0->.93#):

#W=2(.93)^2-2(0)^2#

#W=1.7J#

Note: units of cm were converted to m. One joule, the unit of work, is equivalent to #(kg*m^2)/s^2#. Thus, units of meters are required. This is easier to see in the first example calculation.

Let the actual Height of the tank be #H# and its uniform cross sectional area be #A#.

Now let us consider that at certain moment the height of water level be #h# and the velocity of water emerging through the orifice of cross sectional area #a# at the bottom of the tank be #v#.As the surface of water and the orifice are in open atmosphere, then by Bernoulli's theorem we have

#v=sqrt(2gh)#

Let #dh# represents decrease in water level during infinitesimally small time interval #dt# when the water level is at height #h#. So the rate of decrease in volume of water will be #-A(dh)/(dt)#.

Again the rate of flow of water at this moment through the orifice is given by #vxxa=asqrt(2gh)#.

These two rate must be same by the principle of cotinuity.

Hence #asqrt(2gh)=-A(dh)/(dt)#

#=>dt=-A/(asqrt(2g))*h^(-1/2)dh#

If the tank is filled to the brim then height of water level will be #H# and time required to empty the tank T can be obtained by integrating the above relation.

#T=int_0^Tdt=-A/(asqrt(2g))*int_H^0h^(-1/2)dh#

#T=-A/(asqrt(2g))[h^(1/2)/(1/2)]_H^0#

#T=(A/a*sqrt(2/g))H^(1/2)#

If the tank be half filled with water and the time to empty it be #T'# then

#T'=(A/a*sqrt(2/g))(H/2)^(1/2)#

So #(T')/T=1/sqrt2#

#T'=T/sqrt2#

Answer:

The maximum height of the ball is #~~11m#.

Explanation:

Ignoring air resistance, friction, etc., this problem can be solved with energy conservation.

#E_(mech)=K+U#

Where #K# is kinetic energy and #U# is potential energy. Thus, when energy is conserved in a system it should follow that

#E_i=E_f#

#K_i+U_i=K_f+U_f#

From this relationship we can see that if potential energy, for example, were to decrease, kinetic energy would have to increase in order for energy to be conserved. The opposite is also true. The idea is that energy is being transformed between two forms (potential and kinetic) rather than any of it being lost of gained. When we have to include friction in problems, for example, energy is not conserved, because some of it is lost to heat (#E_(th)#).

Initially, assuming the spring is compressed to point where it's height is insignificant (so that we need not worry about gravitational potential energy of the ball beforehand), all energy is stored as spring potential energy in the spring. Because nothing is moving prior to the launch, there is no kinetic energy. Therefore, we have only potential energy to begin with.

#U_(sp)=U_f+K_f#

After the launch, as a the spring decompresses, the energy stored within it as spring potential energy is transferred to the ball, which is observed as the ball shoots upward. The spring potential energy has been transformed into kinetic energy. However, as the ball rises, it is gaining altitude, and therefore gaining gravitational potential energy. Because the potential energy of the ball increases, its kinetic energy must decrease, and we observe this in the form of the ball stopping at some point in the air before falling back to the earth. If not, it would continue to fly upwards forever!

Because we are concerned with the maximum altitude of the ball, what we're really asking is at what point all of the potential energy from the spring has been transferred into gravitational potential energy, which is given by #U_g=mgh#. When the ball reaches its maximum altitude and pauses momentarily before falling, it has no kinetic energy (remember projectile motion?), and therefore we ultimately have only gravitational potential energy.

#U_(sp)=U_g#

Spring potential energy, #U_(sp)# is given by #1/2k(Δs)^2#.

#1/2k(Δs)^2=mgh#

( tl;dr ) Rearranging to solve for #h#,

#h=(1/2k(Δs)^2)/(mg)#

Using our known values for mass, the spring constant, and the compression of the spring, we can now calculate a value for #h#.

#h=(1/2(18N/m)(4/3m)^2)/(0.150kg*9.8m/s^2)#

#h=10.88...m#

#h≈11m#

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To solve this problem using the knowledge of physics we assume that

  1. the bungee rope behaves as an ideal spring and Hook;s law is obeyed here.

  2. jumper just steps of bridge and falls straight downward.

  3. If the height of the jumper is neglected comparing with the length of the rope, the position of the jumper may be considered as a point.

  4. The loss in gravitational potential energy of the jumper provides the gain in potential energy of the stretched rope at the first lowest position of the jumper, ignoring dissipation of energy during elongation of rope.

Now if the elongation of rope be #x# m for first downward trip then gain in PE of the rope will be
#E_"rope"=1/2xxkx^2#

where

#k->"force constant of the rope "=160Nm^(-1)#

Again the loss PE of jumper

#E_"jumper"=mxxgxxh#

where

#m->"mass of the jumper"=61kg#

#g->"acceleration due to gravity"=10ms^-2#

#h-> "fall of jumper" #

Here

#h="length of relaxed rope" + "elongation of rope"=(20+x)m#

So by conservation of energy

#1/2xxkx^2=mxxgxxh#

#=>1/2xx160x^2=61xx10xx(20+x)#

#=>8x^2=61xx(20+x)#

#=>8x^2-61x-1220=0#

#=>x=1/16(61+sqrt(61^2+4xx8xx1220))~~16.73m#

The jumper jumps from 44m above the water level of the river . If we add the length (20m) of the relaxed rope with the elongation the total elongated length becomes #16.73+20=36.73m#. This length being less than 44m, the jumper will not hit water..

At this stage the upward force on jumper due to elastic property of the rope #Fuarr=kx=160xx16.73N#
and downward gravitational pull #Fdarr=61xx10N#
So net upward pull on the jumper
#F_"net"uarr=Fuarr-Fdarr=160xx16.73N-61xx10N=2066.8N#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let at the final position when oscillation stops the net elongation be #y# m

At this stage the gravitational pull is the cause of stretching and by Hook's law we can write

#kxxy=mg#

#=>y=(mg)/k=(61xx10)/160~~3.81m#
Adding the length of the relaxed rope with it the lowest final position of the jumper will be #20+3.81)=23.81m# .

At this position the net force on the jumper will be Zero as he is at rest.

Answer:

It can be...

Explanation:

Because vector multiplication is not generally arithmetic, but it can be.

A simple example in two dimensions would be if you treat vectors as Complex numbers and define a multiplication #ox# as complex number multiplication:

#[a, b] ox [c, d] = [ac-bd, ac+bd]#

Then there is a corresponding division of vectors:

#[a, b] -: [c, d] = [a, b] ox [c/(c^2+d^2), -d/(c^2+d^2)]#

A more advanced example - useful in mechanics - is the quaternions. Hamilton's quaternions form a 4 dimensional vector space over the real numbers with a natural (though non-commutative) definition of multiplication that makes them into a division algebra, with a natural definition of division.

So treating four dimensional vectors as quaternions, we would define multiplication as:

#[a_1, b_1, c_1, d_1] ox [a_2, b_2, c_2, d_2]#

#=[a_1a_2-b_1b_2-c_1c_2-d_1d_2,#
#color(white)(0000) a_1b_2+b_1a_2+c_1d_2-d_1c_2,#
#color(white)(0000) a_1c_2-b_1d_2+c_1a_2+d_1b_2,#
#color(white)(0000)a_1d_2+b_1c_2-c_1b_2+d_1a_2]#

If this looks a bit like the expansion of matrix multiplication it is no coincidence. Quaternions can be represented by corresponding #4xx4# real matrices of the form:

#((a, -b, -c, -d), (b, a, -d, c), (c, d, a, -b), (d, -c, b, a))#

Then division is basically multiplication by the inverse matrix.

For a very interesting related talk see:

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