Which hemisphere is experiencing summer in july?

A circular 10-turn coil with a radius of 5.0 cm carries a current of 5 A. It lies in the xy plane in a uniform magnetic field =

Tju [1.3M]

Answer:

U = – 0.12J

Explanation:

Given N = 10 turns, I = 5A, r = 5×10-²m

B^ = 0.05 T iˆ+ 0.3 T kˆ

Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T

Area = πr² = π(5×10-²)² = 7.85×10-³m²

Magnetic moment μ = NIA

μ = 10×5×7.85×10-³ = 0.3925Am²

U = -μ•B = –0.3925×0.304 = –0.12J

The sign is negative because the magnetic moment is aligned with the magnetic field.

3 0

3 years ago

Everyone's favorite flying sport disk can be approximated as the combination of a thin outer hoop and a uniform disk, both of di

Umnica [9.8K]

Answer:

The length of the boomerang is 0.364 m

Explanation:

The moment of inertia is:

$I=\frac{1}{2} m_{d} r^{2} +m_{h} r^{2}$

Where

md = 0.05 kg

mh = 0.12 kg

r = d/2 = 0.273/2 = 0.1365 m

$I=\frac{0.05*(0.1365)^{2} }{2} +(0.12)*(0.1365)^{2} =2.7x10^{-3} kgm^{2}$

The length of the boomerang is:

$L_{b} =\sqrt{\frac{12I}{m} } =\sqrt{\frac{12*2.7x10^{-3} }{0.245} } =0.364m$

5 0

3 years ago

suppose a child walks from the outer edge of a rotating Merry-Go-Round to the inside does angular velocity of the Merry-Go-Round

Akimi4 [234]

Okay here initially child walks on the outer edge of the disc and then started to move inside

So here as the child and Merry go round is an isolated system so there is no external Torque on this system from outside

As here we can see there is external force acting on this system by the hinge of the Merry go round as well as due to gravity so we can not use momentum conservation to solve such type of questions.

But as we can say that there is no external torque on this system about the hinge point so we will use conservation of angular momentum for this system

Here as we know that

$\tau = \frac{dL}{dt}$

where L = angular momentum

since here torque is ZERO

$0 = \frac{dL}{dt}$

L = constant

so here we can write initial angular momentum of the system as

$L = (I_1 + I_2)*\omega$

here we know that

$I_1$ = moment of inertia of merry go round

$I_2$ = moment of inertia of child

so here we can say

$(I_1 + I_2)* \omega_1 = (I_1 + I_2')\omega_2$

so here as the child moves from edge to inside the disc it moment of inertia will decrease because as we know that moment of inertia of child is given as

$I_2 = mr^2$