Answer:
a) x = (0.0114 ± 0.0001) in
, b) the number of decks is 5
Explanation:
a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated
x = d / 52
x = 0.590 / 52
x = 0.011346 in
Let's look for uncertainty
Δx = dx /dd Δd
Δx = 1/52 Δd
Δx = 1/52 0.005
Δx = 0.0001 in
The result of the calculation is
x = (0.0114 ± 0.0001) in
b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is
#_cards = n 52
The formula for thickness is
x = d / n 52
Uncertainty
Δx = 1 / n 52 Δd
n = 1/52 Δd / Δx
n = 1/52 0.005 / 0.00002
n = 4.8
Since the number of decks must be an integer the number of decks is 5
Answer:
the object will travel 0.66 meters before to stop.
Explanation:
Using the energy conservation theorem:
![E_i+K_i+W_f=K_f+U_f](https://tex.z-dn.net/?f=E_i%2BK_i%2BW_f%3DK_f%2BU_f)
The work done by the friction force is given by:
![W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]](https://tex.z-dn.net/?f=W_f%3DF_f%2Ad%5C%5CW_f%3D%5C%C2%B5%2Am%2Ag%2Ad%5C%5CW_f%3D0.35%2A4%2A9.81%2Ad%5C%5CW_f%3D13.7d%5BJ%5D)
so:
![\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D1800%2A%2810%2A10%5E%7B-2%7D%29%2B0-13.7d%3D0%2B0%5C%5Cd%3D0.66m)
The answer is C. in sort of a way. You can't technically see black matter. However, it is holding the galaxies together.
Answer:
The tension is ![T= \frac{11}{2\sqrt{3} } Mg](https://tex.z-dn.net/?f=T%3D%20%5Cfrac%7B11%7D%7B2%5Csqrt%7B3%7D%20%7D%20Mg)
The horizontal force provided by hinge ![Fx= \frac{11}{4\sqrt{3} } Mg](https://tex.z-dn.net/?f=Fx%3D%20%5Cfrac%7B11%7D%7B4%5Csqrt%7B3%7D%20%7D%20Mg)
Explanation:
From the question we are told that
The mass of the beam is
The length of the beam is ![l = L](https://tex.z-dn.net/?f=l%20%3D%20L)
The hanging mass is ![m_h = 3M](https://tex.z-dn.net/?f=m_h%20%3D%203M)
The length of the hannging mass is ![l_h = \frac{3}{4} l](https://tex.z-dn.net/?f=l_h%20%3D%20%5Cfrac%7B3%7D%7B4%7D%20l)
The angle the cable makes with the wall is ![\theta = 60^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2060%5Eo)
The free body diagram of this setup is shown on the first uploaded image
The force
are the forces experienced by the beam due to the hinges
Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero
So
![\sum F =0](https://tex.z-dn.net/?f=%5Csum%20F%20%3D0)
Now about the x-axis the moment is
![F_x -T cos \theta = 0](https://tex.z-dn.net/?f=F_x%20-T%20cos%20%5Ctheta%20%20%3D%200)
=> ![F_x = Tcos \theta](https://tex.z-dn.net/?f=F_x%20%3D%20Tcos%20%5Ctheta)
Substituting values
![F_x =T cos (60)](https://tex.z-dn.net/?f=F_x%20%3DT%20cos%20%2860%29)
![F_x= \frac{T}{2} ---(1)](https://tex.z-dn.net/?f=F_x%3D%20%5Cfrac%7BT%7D%7B2%7D%20---%281%29)
Now about the y-axis the moment is
![F_y + Tsin \theta = M *g + 3M *g ----(2)](https://tex.z-dn.net/?f=F_y%20%20%2B%20Tsin%20%5Ctheta%20%20%3D%20M%20%2Ag%20%2B%203M%20%2Ag%20----%282%29)
Now the torque on the system is zero because their is no rotation
So the torque above point 0 is
![M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0](https://tex.z-dn.net/?f=M%2A%20g%20%2A%20%5Cfrac%7BL%7D%7B2%7D%20%20%2B%203M%20%2A%20g%20%5Cfrac%7B3L%7D%7B2%7D%20-%20T%20sin%2860%29%20%2A%20L%20%3D%200)
![\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0](https://tex.z-dn.net/?f=%5Cfrac%7BMg%7D%7B2%7D%20%2B%20%5Cfrac%7B9%20Mg%7D%7B4%7D%20-%20%20T%20%2A%20%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B2%7D%20%20%20%20%3D%200)
![\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7B2Mg%20%2B%209Mg%7D%7B4%7D%20%3D%20T%20%2A%20%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B2%7D)
![T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B11Mg%7D%7B4%7D%20%2A%20%5Cfrac%7B2%7D%7B%5Csqrt%7B3%7D%20%7D)
![T= \frac{11}{2\sqrt{3} } Mg](https://tex.z-dn.net/?f=T%3D%20%5Cfrac%7B11%7D%7B2%5Csqrt%7B3%7D%20%7D%20Mg)
The horizontal force provided by the hinge is
![F_x= \frac{T}{2} ---(1)](https://tex.z-dn.net/?f=F_x%3D%20%5Cfrac%7BT%7D%7B2%7D%20---%281%29)
Now substituting for T
![F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}](https://tex.z-dn.net/?f=F_%7Bx%7D%20%3D%20%5Cfrac%7B11%7D%7B2%5Csqrt%7B3%7D%20%7D%20%2A%20%5Cfrac%7B1%7D%7B2%7D)
![Fx= \frac{11}{4\sqrt{3} } Mg](https://tex.z-dn.net/?f=Fx%3D%20%5Cfrac%7B11%7D%7B4%5Csqrt%7B3%7D%20%7D%20Mg)