Answer:
it is light
Explanation:
the arrow that says light is on the glass it must be near from tungsten
Answer:
The force on each wire is
![T_1 = 12.5 \ N](https://tex.z-dn.net/?f=T_1%20%20%3D%2012.5%20%5C%20N%20%20)
![T_2 = 25 \ N](https://tex.z-dn.net/?f=%20%20T_2%20%20%3D%20%20%2025%20%5C%20%20N%20%20)
![T_3 = 50 \ N](https://tex.z-dn.net/?f=%20%20T_3%20%20%3D%20%2050%20%20%5C%20%20N%20%20)
Explanation:
From the question we are told that
The acceleration at which the elevator will stop is ![a = \frac{g}{4}](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bg%7D%7B4%7D)
The weight of each section of the wire is ![W = \ 10 \ N](https://tex.z-dn.net/?f=W%20%3D%20%20%5C%2010%20%5C%20N)
Generally
here
are weight at each section
Generally considering the first section, the force acting along the y-axis is mathematically represented as
![\sum F_y_1 = T_1 - W_1 = m * a](https://tex.z-dn.net/?f=%5Csum%20F_y_1%20%3D%20%20T_1%20-%20W_1%20%3D%20%20m%20%2A%20%20a)
Here
represents the tension on the wire at the first section while
represents the weight of the lamp at the first section
So
![T_1 - 10 = m * \frac{g}{4}](https://tex.z-dn.net/?f=T_1%20%20-%2010%20%3D%20m%20%2A%20%20%5Cfrac%7Bg%7D%7B4%7D)
=> ![T_1 - 10 = \frac{W_1}{4}](https://tex.z-dn.net/?f=T_1%20%20-%2010%20%3D%20%20%5Cfrac%7BW_1%7D%7B4%7D)
=> ![T_1 - 10 = \frac{10}{4}](https://tex.z-dn.net/?f=T_1%20%20-%2010%20%3D%20%20%5Cfrac%7B10%7D%7B4%7D)
=> ![T_1 = 12.5 \ N](https://tex.z-dn.net/?f=T_1%20%20%3D%2012.5%20%5C%20N%20%20)
Generally considering the second section, the force acting along the y-axis is mathematically represented as
![\sum F_y_2 = T_2 -T_1- W_2 = m * a](https://tex.z-dn.net/?f=%5Csum%20F_y_2%20%3D%20%20T_2%20-T_1-%20W_2%20%3D%20%20m%20%2A%20%20a)
=> ![T_2 - T_1- 10 = m * \frac{g}{4}](https://tex.z-dn.net/?f=%20%20T_2%20-%20T_1-%2010%20%3D%20%20m%20%2A%20%20%5Cfrac%7Bg%7D%7B4%7D)
=> ![T_2 - 12.5- 10 = \frac{W_2}{4}](https://tex.z-dn.net/?f=%20%20T_2%20-%2012.5-%2010%20%3D%20%20%20%20%5Cfrac%7BW_2%7D%7B4%7D)
=> ![T_2- 12.5- 10 = \frac{10}{4}](https://tex.z-dn.net/?f=%20%20T_2-%2012.5-%2010%20%3D%20%20%20%20%5Cfrac%7B10%7D%7B4%7D)
=> ![T_2 = 25 \ N](https://tex.z-dn.net/?f=%20%20T_2%20%20%3D%20%20%2025%20%5C%20%20N%20%20)
Generally considering the third section, the force acting along the y-axis is mathematically represented as
![\sum F_y_3 = T_3- T_2 -T_1- W_3 = m * a](https://tex.z-dn.net/?f=%5Csum%20F_y_3%20%3D%20T_3-%20T_2%20-T_1-%20W_3%20%3D%20%20m%20%2A%20%20a)
![T_3 - T_2 - T_1- 10 = m * \frac{g}{4}](https://tex.z-dn.net/?f=%20%20T_3%20-%20T_2%20-%20T_1-%2010%20%3D%20%20m%20%2A%20%20%5Cfrac%7Bg%7D%7B4%7D)
![T_3 - 25 - 12.5- 10 = \frac{W_2}{4}](https://tex.z-dn.net/?f=%20%20T_3%20-%2025%20-%2012.5-%2010%20%3D%20%20%20%20%5Cfrac%7BW_2%7D%7B4%7D)
![T_3 - 25 - 12.5- 10 = \frac{10}{4}](https://tex.z-dn.net/?f=%20%20T_3%20-%2025%20-%2012.5-%2010%20%3D%20%20%20%20%5Cfrac%7B10%7D%7B4%7D)
![T_3 = 50 \ N](https://tex.z-dn.net/?f=%20%20T_3%20%20%3D%20%2050%20%20%5C%20%20N%20%20)
Answer:
Equal
Explanation:
Recall that KE = 1/2(m)(v²) and PE = mgh, where m is mass, g is gravity, v is velocity, and h is height.
When the bowling ball is first dropped, it has a maximum potential energy but minimum kinetic energy. The height is max, so the potential energy will be greatest. Velocity is 0, so kinetic energy will be 0.
When the bowling ball is half way through its fall, the height will be half the initial height and the velocity will be half of the final velocity.
When the bowling ball is at the bottom and reaches the ground, the height is 0 so potential energy is 0 while the kinetic energy is max because velocity is the greatest.
Answer:
The minimum acceleration will be
Explanation:
We have given that coefficient of static friction ![\mu _s=0.860](https://tex.z-dn.net/?f=%5Cmu%20_s%3D0.860)
Frictional force is given by ![F=\mu_s N=\mu _sma](https://tex.z-dn.net/?f=F%3D%5Cmu_s%20N%3D%5Cmu%20_sma)
Weight is given by ![W=mg](https://tex.z-dn.net/?f=W%3Dmg)
For train to allow the bat to remain in place ![mg=\mu _sma](https://tex.z-dn.net/?f=mg%3D%5Cmu%20_sma)
![a=\frac{g}{\mu _s}=\frac{9.81}{0.860}=11.40m/sec^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bg%7D%7B%5Cmu%20_s%7D%3D%5Cfrac%7B9.81%7D%7B0.860%7D%3D11.40m%2Fsec%5E2)
So the minimum acceleration will be ![11.40m/sec^2](https://tex.z-dn.net/?f=11.40m%2Fsec%5E2)
Answer:
let us use the expression v = ./2gS in both the cases.
a) In the first case g = 9.8 m/s^2 and S = 10.3 m.
So v = 14.21 m/s
b) in the second case, g = 1.64 m/s^2 and S = 4.2 m = 3.71 m/s
00
Explanation: