Answer:
The torque on the child is now the same, τ.
Explanation:
- It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
- In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
- The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
- When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:
- When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:
- Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:
τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.
Answer:
Finally current will be
i = 0.35 A
Explanation:
As we know that power of the bulb is given by the formula
now we have
R = 240 ohm
so we have
now the current in the bulb is given as
now when length of the filament is double
so the resistance of the wire also gets double
so we have
now the current in the bulb is given as
Answer:
a. N = 2.49W b. 0.40
Explanation:
a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?
Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s
Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m
The rider's weight W = mg = 9.8m
The ratio of the normal force to the rider's weight is
N/W = 24.43m/9.8m = 2.49
So the normal force expressed in term's of the rider's weight is
N = 2.49W
b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?
The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.
Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.
So, the normal force equals
N = F/μ = W/μ = mg/μ = mrω²
μ = mg/mrω²
= W/N
= 9.8m/24.43m
= 0.40
Answer:
a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m
Explanation:
The diffraction network is described by the expression
d sin θ= m λ
Where m corresponds to the diffraction order
Let's use trigonometry to find the breast
tan θ = y / L
The diffraction spectrum is measured at very small angles, therefore
tan θ = sin θ / cos θ = sin θ
We replace
d y / L = m λ
Let's place in the first order m = 1
Let's look for the separation of the lines (d)
d = λ L / y
d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²
d = 332.33 10⁻⁹ m
Now we can look for the wavelength of the other line
λ = d y / L
λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²
λ = 189.43 10⁻⁹ m
Part B
The compound wavelength B
λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²
λ = 269.19 10⁻⁹ m
Answer:
A). A virtual image cannot be formed on a screen.
Explanation:
A virtual image can not be formed on a screen.
For image:
1.A virtual image can be viewed by the unaided eye.
2. A real image must be erect or maybe inverted.
3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.
4.A virtual image can be photographed.
So the option A is correct.
