Since it is restricted at both ends, λ/2 = length of string
λ/2 = 1.5m
λ = 1.5*2 = 3m
Answer:
v = 21 m / s
Explanation:
We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data
v = v₀ + a t
the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s
a = (v-v₀) / t
a = (11 - 6) / 8
a = 0.625 m / s²
when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s
v = v₀ + a t
v = 11 + 0.625 16
v = 21 m / s
Answer:
The current in second wire is 5.0 A.
(B) is correct option.
Explanation:
Given that,
Current in first wire = 3.7 A
Distance = 8.0 cm
We need to calculate the magnetic field due to the current carrying wire
Using formula of magnetic field

Where, I = current
r = distance
Put the value into the formula
For first wire
...(I)
For second wire,
The distance is 8-3.7 = 4.3 cm
...(II)
The magnetic field in both the wires,
From equation (I) and (II)



Hence, The current in second wire is 5.0 A.
<h2>
Answer:20.97g N,32.63g N</h2>
Explanation:
We consider the forces at the knot.
The vertical forces are
is the vertical component of tension
at the knot.
is the weight of the mass
acting downwards.
The horizontal forces are
is the tension in the rope acting left.
is the horizontal component of tension
acting towards right.
Since the knot has no mass,it is always in equilibrium.
So,the sum of forces acting on it will be zero.
Balancing vertical forces gives,


=
Balancing horizontal forces gives,


