The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
Explanation:
Consider the vertical motion of ball,
We have equation of motion v = u + at
Initial velocity, u = u sin θ
Final velocity, v = 0 m/s
Acceleration = -g
Substituting
v = u + at
0 = u sin θ - g t
![t=\frac{usin\theta }{g}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Busin%5Ctheta%20%7D%7Bg%7D)
This is the time of flight.
Consider the horizontal motion of ball,
Initial velocity, u = u cos θ
Acceleration, a =0 m/s²
Time,
Substituting
s = ut + 0.5 at²
![s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}](https://tex.z-dn.net/?f=s%3Ducos%5Ctheta%20%5Ctimes%20%5Cfrac%7Busin%5Ctheta%20%7D%7Bg%7D%2B0.5%5Ctimes%200%5Ctimes%20%28%5Cfrac%7Busin%5Ctheta%20%7D%7Bg%7D%29%5E2%5C%5C%5C%5Cs%3D%5Cfrac%7Bu%5E2sin%5Ctheta%20cos%5Ctheta%7D%7Bg%7D%5C%5C%5C%5Cs%3D%5Cfrac%7Bu%5E2sin2%5Ctheta%7D%7B2g%7D)
This is the range.
In this problem
u = 30 m/s
g = 9.81 m/s²
θ = 45° - For maximum range
Substituting
![s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B30%5E2%5Ctimes%20sin%282%5Ctimes%2045%29%7D%7B2%5Ctimes%209.81%7D%3D45.87m)
Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.
So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
We can use the equation for kinetic energy, K=1/2mv².
Your given variables are already in the correct units, so we can just plug in the variables and solve for v.
K = 1/2mv²
16 = 1/2(2)v²
16 = (1)v²
√16 = v
v = 4 m/s
Therefore, the velocity of a 2 kg mass with 16 J of kinetic energy is 4 m/s.
Hope this is helpful!
You can make sure there's no change in volume by keeping
your gas in a sealed jar with no leaks. Then you can play with
the temperature and the pressure all you want, and you'll know
that the volume is constant.
For 'ideal' gases,
(pressure) times (volume) is proportional to (temperature).
And if volume is constant, then
(pressure) is proportional to (temperature) .
So if you increase the temperature from 110K to 235K,
the pressure increases to (235/110) of where it started.
(400 kPa) x (235/110) = 854.55 kPa. (rounded)
Obviously, choice-b is the right one, but
I don't know where the .46 came from.
ಠ_ಠ Hey, hang on.. you might've made a discovery. Nobody has tested it so how do we know? ಠ_ಠ
Answer:
![0.53\times 10^{-30}kgms^{-1}](https://tex.z-dn.net/?f=0.53%5Ctimes%2010%5E%7B-30%7Dkgms%5E%7B-1%7D)
Explanation:
Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,
![\Delta x\Delta p=\frac{h}{4\pi }](https://tex.z-dn.net/?f=%5CDelta%20x%5CDelta%20p%3D%5Cfrac%7Bh%7D%7B4%5Cpi%20%7D)
Given that the uncertainty in x is 0.1 mm.
Therefore,
![\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}](https://tex.z-dn.net/?f=%5CDelta%20p%3D%5Cfrac%7B6.626%5Ctimes%2010%5E%7B-34%7D%20%7D%7B4%5Ctimes%203.14%5Ctimes%201%5Ctimes%2010%5E%7B-4%7Dm%20%7D%5C%5C%5CDelta%20p%3D0.53%5Ctimes%2010%5E%7B-30%7Dkgms%5E%7B-1%7D)
Therefore, uncertainty in the transverse momentum of photon is ![0.53\times 10^{-30}kgms^{-1}](https://tex.z-dn.net/?f=0.53%5Ctimes%2010%5E%7B-30%7Dkgms%5E%7B-1%7D)