An electromagnet is made from a coil of wire which acts as a magnet when an electric current passes through it. Often an electromagnet is wrapped around a core of ferromagnetic material like steel, which enhances the magnetic field produced by the coil.
Answer:
A) mr = 100 kg
B) Fr = 210N
C) Ft = -199.5N
Explanation:
By conservation of the momentum:
mt*Vo = (mr + mt) * Vf Solving for mr:
mr = mt*Vo / Vf - mt = 100 kg
The average force on the receiver:
mr *(Vf - 0) = Fr * Δt Solving for Fr:
Fr = 210 N
The average force on the tackler:
mt * (Vf - Vo) = Ft * Δt Solving for Ft:
Ft = -199.5 N
Answer:
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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,