Ediment deposited where a river flows into an ocean or lake

Is it possible to orient a current loop in a uniform magnetic field such that the loop does not tend to rotate? Explain.

vazorg [7]

Yes, it possible to orient a current loop in a uniform magnetic field such that the loop will not tend to rotate.

The forces on opposing sides of the loop would be equal in magnitude and opposite in direction if the magnetic field is pointed perpendicular to the plane of the loop; yet, there will be no net torque produced on the loop. In a constant magnetic field, there is no torque experienced by a current loop. A current carrying loop in a homogeneous magnetic field will never exert any net magnetic force. In a constant magnetic field, there is no torque experienced by a current loop. Due to the zero torque on the loop in this position, the plane of the loop is parallel to the direction of the magnetic field.

To know more about uniform magnetic field refer to brainly.com/question/25655915

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5 0

2 years ago

Which is an example of kinetic energy? A. a stretched rubber band B. wind C. water in a reservoir D. natural gas E. an object su

daser333 [38]

Kinetic energy is energy of motion.

In the cases of a stretched rubber band, water in a reservoir, natural gas, or an object suspended above the ground, everything is just laying there, and nothing is moving. There's nothing there that has kinetic energy.

If there's any wind, then air is moving. The moving air has kinetic energy.

6 0

4 years ago

How many planet pinions are used in a planetary gear set? quizlit?

timama [110]

Im thinking maybe..6 im just thinking

6 0

3 years ago

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 58.2 cm, and the density of iro

vredina [299]

Answer:

0.556m

Explanation:

unit conversion

58.2 cm = 0.582 m -> the outer radius is 0.582/2 = 0.291 m

$7.87 g/cm^33 = 7.87 g/cm^33 * 1/1000 (kg/g) * 100^3 cm^3/m^3 = 7870 kg/m^3$

Let r be the inner diameter we can find the volume of the hollow part of the sphere using the following formula

$V_h = \frac{4}{3}\pi r^3$

The volume of the iron shell is the volume of the whole sphere subtracted by volume of the hollow part

$V_s = V - V_h = \frac{4}{3}\pi 0.291^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.0246 - r^3)$

Let water density $\rho_w = 1000 kg/m^3$ , then the buoyant force is the weight of water displaced by the shell

$F_b = g\rho_w V = g1000\frac{4}{3}\pi(0.291^3) = g1000\frac{4}{3}\pi(0.0246)$

For the shell to almost completely submerged in water, its buoyant force equal to the weight of the shell

$F_b = W_s$

$g1000\frac{4}{3}\pi(0.0246) = g\rho_sV_s$

$g1000\frac{4}{3}\pi(0.0246) = g7870\frac{4}{3}\pi(0.0246 - r^3)$

$1000*0.0246 = 7870(0.0246 - r^3)$

$24.6 = 193.602 - 7870r^3$

$r^3 = (193.602 - 24.6)/7870 = 0.02147$

$r = \sqrt[3]{0.02147} = 0.278 m$

So the inner diameter is 0.278*2 = 0.556 m

3 0

3 years ago

A point charge is placed at the center of a spherical Gaussian surface. Is changed (a) if the surface is replaced by a cube of t

Jlenok [28]

Explanation:

Electric flux in enclosed surface depends on the total charge inside the surface

It is independent of shape of the Gaussian surface as well as it is independent of position of charge inside the surface

so we have

(a) if the surface is replaced by a cube of the same volume?

No change in the flux as net charge inside the Gaussian surface remains unchanged

(b) if the sphere is replaced by a cube of one-tenth the volume?

No change in the flux as net charge inside the Gaussian surface remains unchanged

(c) if the charge is moved off center in the original sphere, still remaining inside?

No change in the flux as net charge inside the Gaussian surface remains unchanged

(d) if the charge is moved just outside the original sphere?

Flux will change to ZERO because there is no charge inside the surface

(e) if a second charge is placed near, and outside, the original sphere?

No change in flux as the second charge is outside the surface

(f) if a second charge is placed inside the Gaussian surface?

Flux will change as net charge inside the surface will change

3 0

3 years ago