4 US regions where major aquifers can be found

A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the gro

grin007 [14]

Answer:

torque is 1.7 * $10^{-2}$ Nm

Explanation:

Given data

turns n = 1000 turns

radius r = 12 cm

current I = 15A

magnitude B = 5.8 x 10^-5 T

angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

torque = 1000* 15* π(0.12)² *5.8 x 10-5 * sin25

torque = 0.0166 N m

torque is 1.7 * $10^{-2}$ Nm

5 0

3 years ago

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

konstantin123 [22]

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

$L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}$

Where

$I_{i}\ and \ \omega_{i}$ initial moment of inertia and angular velocity

$I_{f}\ and \ \omega_{f}$ is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ is the satellite mass

r is the radus of the sphere

Substititute 1900kg for m and 4.6m for r

$I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}$

The final moment of inertia of the satellite about the centre of mass

$I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}$

Where $m_{x}$ is the antenna's mass and

I is the length of the antenna

$I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}$

So, the Final rotation rate of the satellite is:

$I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

5 0

3 years ago

In _______ atomic model, negative electrons orbit the positively charged nucleus.

Shalnov [3]

In (A. Dalton's) atomic model, negative electrons orbit the positively charged nucleus.

8 0

3 years ago

Read 2 more answers

With Christmas lights wired in series, if one light goes out, what happens to the rest?

daser333 [38]

Answer:

The ones that are after the light that went out are also out.

Explanation:

3 0

2 years ago

Read 2 more answers

A long, East-West-oriented power cable carrying an

Alla [95]

Answer:

200A

Explanation:

Given that

the distance between earth surface and power cable d = 8m

when the current is flowing through cable , the magnitude flux density at the surface is 15μT

when the current flow throught is zero the magnitude flux density at the surface is 20μT

The change in flux density due to the current flowing in the power cable is

B = 20μT - 15μT

B =5μT -----(1)

The expression of magnitude flux density produced by the current carrying cable is

$B=\frac{\mu_0I}{2\pi d}$ -----(2)

Substitute the value of flux density

B from eqn 1 and eqn 2

$\frac{\mu_0I}{2\pi d}=5\times 10^-^6\\\\\frac{(4\pi \times 10^-^7)I}{2 \pi (8)} =5\times 10^-^6\\\\I=200A$

Therefore, the magnitude of current I is 200A

8 0

3 years ago