Answer:
Resistance = 3.35*
Ω
Explanation:
Since resistance R = ρ
whereas 
resistivity is given for two ends. At the left end resistivity is 
whereas x at the left end will be 0 as distance is zero. Thus
At the right end x will be equal to the length of the rod, so 
Thus resistance will be R = ρ
where A = π 
so,
Answer:
|X| = 72 cm
Explanation:
We need to find the magnitude of vector X.
Vector Y = 21 cm
Vector Z = 75 cm
We know that, the magnitude of resultant vector is given by :
So, the value of magnitude of vector X is 72 cm
Then the workers that made poison for rats would have to throw all of the poison in the garbage, sad, I know. Oh, if rats was eliminated from the environment, then cats that LOVE rats wont be poisoned by the rats that EAT the poison :). Yes, one of my mama cats died );
Answer:
a. a = 
b. T = -0.81 N
Explanation:
Given,
- weight of the lighter block =
- weight of the heavier block =
- inclination angle =
- coefficient of kinetic friction between the lighter block and the surface =
- coefficient of kinetic friction between the heavier block and the surface =
- friction force on the lighter block =
- friction force on the heavier block =
Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.
From the f.b.d. of the lighter block,
From the f.b.d. of the heavier block,
From eqn (1) and (2), we get,
![\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\](https://tex.z-dn.net/?f=%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7Bg%28w_1sin%5Ctheta%5C%20-%5C%20%5Cmu_1w_1cos%5Ctheta%5C%20%2B%5C%20w_2sin%5Ctheta%5C%20%2B%5C%20%5Cmu_2w_2cos%5Ctheta%29%7D%7Bw_1%5C%20%2B%5C%20w_2%7D%5C%5C%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7Bg%5Bsin%5Ctheta%28w_1%5C%20%2B%5C%20w_2%29%5C%20%2B%5C%20cos%5Ctheta%28%5Cmu_2w_2%5C%20-%5C%20%5Cmu_1w_1%29%5D%7D%7Bw_1%5C%20%2B%5C%20w_2%7D%5C%5C)
![\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.](https://tex.z-dn.net/?f=%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7B9.81%5Ctimes%20%5Bsin30%5Eo%5Ctimes%20%283.0%5C%20%2B%5C%207.0%29%5C%20%2B%5C%20cos30%5Eo%5Ctimes%20%280.31%5Ctimes%207.0%5C%20-%5C%200.13%5Ctimes%203.0%29%5D%7D%7B3.0%5C%20%2B%5C%207.0%7D%5C%5C%5CRightarrow%20a%5C%20%3D%5C%206.4%5C%20m%2Fs.)
part (b)
From the eqn (2), we get,
From Newton's law v^2 = u^2 + 2as where a is the acceleration and s is the distance.
But to go any further, we need to know how fast the vehicle is accelerating
From v = u +at
We have a = u/t where the final velocity v = 0
So in one minute acceleration = (35 / 60) / 60 = 0.0097 ms/2. The first
experession in bracket is the initial velocity, u, in metres per seconds.
Hence v^2 = (0.583)^2 + 2 (0.0097)(30)
v^2 = 0.3398 + 0.5826 = 0.9224
v = âš 0.9224 = 0.960m
