Answer:
Explanation:
a ) work done by gravitational force
= mg sinθ ( d + .21)
Potential energy stored in compressed spring
= 1/2 k x²
= .5 x 431 x ( .21 )²
= 9.5
According to conservation of energy
mg sinθ ( d + .21) = 9.5
3.2 x 9.8 x sin 30( d + .21 ) = 9.5
d = 40 cm
b )
As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.
mg sin30 = kx
3.2 x 9.8 x .5 = 431 x
x = 3.63 cm
When there is compression of 3.63 cm in the spring , block will have maximum velocity. there after its speed will start decreasing.
Hello!
Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.
![\Sigma \tau = 0](https://tex.z-dn.net/?f=%5CSigma%20%5Ctau%20%3D%200)
We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)
- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)
Both of these act in opposite directions. Let's use the equation for torque:
![\tau = r \times F](https://tex.z-dn.net/?f=%5Ctau%20%3D%20r%20%5Ctimes%20F)
Doing the summation using their respective lever arms:
![0 = L Tsin\theta - dF_g](https://tex.z-dn.net/?f=0%20%3D%20L%20Tsin%5Ctheta%20%20-%20dF_g)
![dF_g = LTsin\theta](https://tex.z-dn.net/?f=dF_g%20%3D%20LTsin%5Ctheta)
Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:
![tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7BH%7D%7BL%7D%5C%5C%5C%5Ctan%5E%7B-1%7D%28%5Cfrac%7BH%7D%7BL%7D%29%20%3D%20%5Ctheta%5C%5C%5C%5Ctan%5E%7B-1%7D%28%5Cfrac%7B1.70%7D%7B2.200%7D%29%20%3D%2037.69%5Eo)
Now, let's solve for 'T'.
![T = \frac{dMg}{Lsin\theta}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7BdMg%7D%7BLsin%5Ctheta%7D)
Plugging in the values:
![T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B%280.700%29%284.00%29%289.8%29%7D%7B%282.200%29sin%2837.69%29%7D%20%3D%20%5Cboxed%7B20.399%20N%7D)
Answer:
Explanation:
Given
Force ![F=1.6\times 10^{6} N](https://tex.z-dn.net/?f=F%3D1.6%5Ctimes%2010%5E%7B6%7D%20N)
one at an angle of
East of North and another at
West of North
Net Force is in North Direction
![F_{net}=2F\cos 13](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D2F%5Ccos%2013)
Forces in horizontal direction will cancel out each other
thus Work done will be by north direction forces
![W=2F\cdot \cos 30\cdot s](https://tex.z-dn.net/?f=W%3D2F%5Ccdot%20%5Ccos%2030%5Ccdot%20s)
here ![s=0.7 km](https://tex.z-dn.net/?f=s%3D0.7%20km)
![W=2\times 1.6\times 10^{6}\cdot \cos 30\cdot 700](https://tex.z-dn.net/?f=W%3D2%5Ctimes%201.6%5Ctimes%2010%5E%7B6%7D%5Ccdot%20%5Ccos%2030%5Ccdot%20700)
Answer:
Order of 10^(-35) m.
Explanation:
The string theory is a theoretical concept whereby the very small particles of particle physics are replaced by one dimensional objects which are called strings. This theory is also applicable to black hole physics, nuclear physics, cosmology, etc.
Now, according to string theory, six space-time dimensions cannot be measured except as quantum numbers of internal particle properties because they are curled up in size of the order of 10^(-35) m.
This is because the length of the scale is assumed to be on the order of the Planck length, or 10^(−35) meters which is the scale at which the effects of quantum gravity are usually believed to become very significant.
The most dense thing inside the column is the little star. It's pink, and it's resting on the bottom of the cylinder.
The red part under the star is the solid base that the cylinder is standing in. It's probably wood or plastic, it's not free to move, and it's not involved in the experiment.