I was about to say: because people generally get comfortable with
what they think they know, and don't like the discomfort of being told
that they have to change something they're comfortable with.
But then I thought about it a little bit more, and I have a different answer.
"Society" might initially reject a new scientific theory, because 'society'
is totally unequipped to render judgement of any kind regarding any
development in Science.
First of all, 'Society' is a thing that's made of a bunch of people, so it's
inherently unequipped to deal with scientific news. Anything that 'Society'
decides has a lot of the mob psychology in it, and a public opinion poll or
a popularity contest are terrible ways to evaluate a scientific discovery.
Second, let's face it. The main ingredient that comprises 'Society' ... people ...
are generally uneducated, unknowledgeable, unqualified, and clueless in the
substance, the history, and the methods of scientific inquiry and reporting.
There may be very good reasons that some particular a new scientific theory
should be rejected, or at least seriously questioned. But believe me, 'Society'
doesn't have them.
That's pretty much why.
Answer:
![14.98\ \text{kg m/s}](https://tex.z-dn.net/?f=14.98%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
![45.26^{\circ}](https://tex.z-dn.net/?f=45.26%5E%7B%5Ccirc%7D)
Explanation:
= Initial momentum of the pin = 13 kg m/s
= Initial momentum of the ball = 18 kg m/s
= Momentum of the ball after hit
= Angle ball makes with the horizontal after hitting the pin
= Angle the pin makes with the horizotal after getting hit by the ball
Momentum in the x direction
![P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}](https://tex.z-dn.net/?f=P_i%3DP_1%5Ccos55%5E%7B%5Ccirc%7D%2BP_2%5Ccos%5Ctheta%5C%5C%5CRightarrow%20P_2%5Ccos%5Ctheta%3DP_i-P_1%5Ccos55%5E%7B%5Ccirc%7D%5C%5C%5CRightarrow%20P_2%5Ccos%5Ctheta%3D18-13%5Ccos55%5E%7B%5Ccirc%7D%5C%5C%5CRightarrow%20P_2%5Ccos%5Ctheta%3D10.54%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
Momentum in the y direction
![P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}](https://tex.z-dn.net/?f=P_1%5Csin55%3DP_2%5Csin%5Ctheta%5C%5C%5CRightarrow%20P_2%5Csin%5Ctheta%3D13%5Csin55%5E%7B%5Ccirc%7D%5C%5C%5CRightarrow%20P_2%5Csin%5Ctheta%3D10.64%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
![(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}](https://tex.z-dn.net/?f=%28P_2%5Ccos%5Ctheta%29%5E2%2B%28P_2%5Csin%5Ctheta%29%5E2%3DP_2%5E2%5C%5C%5CRightarrow%20P_2%3D%5Csqrt%7B10.54%5E2%2B10.64%5E2%7D%5C%5C%5CRightarrow%20P_2%3D14.98%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
The pin's resultant velocity is ![14.98\ \text{kg m/s}](https://tex.z-dn.net/?f=14.98%5C%20%5Ctext%7Bkg%20m%2Fs%7D)
![P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}](https://tex.z-dn.net/?f=P_2%5Csin%5Ctheta%3D10.64%5C%5C%5CRightarrow%20%5Ctheta%3Dsin%5E%7B-1%7D%5Cdfrac%7B10.64%7D%7B14.98%7D%5C%5C%5CRightarrow%20%5Ctheta%3D45.26%5E%7B%5Ccirc%7D)
The pin's resultant direction is
below the horizontal or to the right.
Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.
The power require to keep the car traveling is 6,666 W.
The power of the engine at the given efficiency is 3,999.6 W.
<h3>What is Instantaneous power?</h3>
This the product of force and velocity of the given object.
The power require to keep the car traveling is calculated as follows;
P = Fv
![P = 300\ N \ \times \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\ P = 300 \ N \times 22.22 \ m/s\\\\ P = 6,666 \ W](https://tex.z-dn.net/?f=P%20%3D%20300%5C%20N%20%5C%20%5Ctimes%20%20%5C%20%5Cfrac%7B80%20%5C%20kmh%5E%7B-1%7D%7D%7B3.6%20%5C%20km%20h%5E%7B-1%7D%2Fm%2Fs%7D%20%5C%5C%5C%5C%0AP%20%3D%20300%20%5C%20N%20%5Ctimes%2022.22%20%5C%20m%2Fs%5C%5C%5C%5C%0AP%20%3D%206%2C666%20%5C%20W)
The power of the engine at the given efficiency is calculated as follows;
![E = \frac{P_{out}}{P _{in}} \times 100\%\\\\ 60\% = \frac{P_{out}}{6,666} \times 100\%\\\\ 0.6 = \frac{P_{out}}{6,666} \\\\ P_{out} = 3,999.6 \ W](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BP_%7Bout%7D%7D%7BP%20_%7Bin%7D%7D%20%5Ctimes%20100%5C%25%5C%5C%5C%5C%0A60%5C%25%20%3D%20%5Cfrac%7BP_%7Bout%7D%7D%7B6%2C666%7D%20%5Ctimes%20100%5C%25%5C%5C%5C%5C%0A0.6%20%3D%20%5Cfrac%7BP_%7Bout%7D%7D%7B6%2C666%7D%20%5C%5C%5C%5C%0AP_%7Bout%7D%20%3D%203%2C999.6%20%5C%20W)
Learn more about efficiency here: brainly.com/question/15418098
Oil is less dense than water so it tends to float on the top of the water. Hope this Helps!