Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T
1 answer:
Answer:
40 N
Explanation:
We are given that
Speed of system is constant
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B
Where =Mass of block B
Substitute the values
For block A
Where Mass of block A
Substitute the values
Hence, the friction force acting on block A=40 N
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Explanation:
The nearest distance is:
(eq. 1)
Where
z = atomic number of gold = 79
e = electron charge = 1.6x10⁻¹⁹C
k = electrostatic constant = 9x10⁹Nm²C²
energy of the particle = 32 MeV = 5.12x10⁻¹²J
At the potential energy is zero, all the energy will be kinetic energy:
Where
m = 4 mp = mass of proton
Replacing in equation 1
Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ =
q₂ =
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀