Can reverse magnetic compulsion be turned in to a form of non-lethal weaponry?

Why does charging by rubbing happens best in dry weather?

navik [9.2K]

Maybe friction..? my best answer.

7 0

4 years ago

A loop of wire is placed in a uniform magnetic field. Let ø be the angle between the direction of b and the normal. At which val

zaharov [31]

Answer:

B. 0 degrees

Explanation:

The magnetic flux through a certain area enclosed by a loop of wire immersed in a region with magnetic field is given by:

$\Phi = B A cos \theta$

where

B is the strength of the magnetic field

A is the area enclosed by the coil

$\theta$ is the angle between the direction of the magnetic field and the direction of the normal to the coil

From the equation above, we see that the magnetic flux is maximum when

$cos \theta=1$

So when the field is parallel to the normal to the coil (so, when it is perpendicular to the coil), therefore when:

$\theta=0^{\circ}$

So, the correct option is

B. 0 degrees

5 0

4 years ago

In a Young’s interference experiment, the two slits are separated by 0.150 mm and the incident light includes two wavelengths: l

dezoksy [38]

Answer:

2.52 × 10⁻² cm

Explanation:

The distance of bright fringe from the center of the screen is given by the formula

$y = \frac{m\lambda D}{d}$

Here, wavelength is λ, Distance of the screen from the slits is D, seperation between the

slits is d.

Separation between the slits, d = 0.15 mm

$= 0.15 * 10^{-3} m$

Distance of the screen from the slits = 1.40 m

We have a wavelength, λ1 = 540 nm

= $540 * 10^{-9} m$

By substituting all these values in the above equation we get

y1 = mλD/d

$y1 = m(540 \times 10^{-9} m)(1.40 m)/(0.15 \times 10^{-3} m)$

$y1 = m(5.04 * 10^{-3} m)$

We have a wavelength, λ2 = 450 nm

= $450 * 10^-9 m$

By substituting all these values in the above equation we get

$y_2 = \frac{m\lambda D}{d}$

$y_2 = m(450 * 10^{-9} m)(1.40 m)/(0.15 * 10^{-3} m)$

$y_2 = m'(4.20 * 10^{-3} m)$

According to the problem, these two distance are coincides with each other.

So,

$y_1 = y_2$

$m(5.04 * 10^{-3} m) = m'(4.20 * 10^{-3} m)$

by testing values, the above equation is satisfied only when, m = 5 and m' = 6

Then from the above we have

y1 = y2 = 0.0252 m

= 2.52 × 10⁻² cm

8 0

3 years ago

Help with physical science please

alex41 [277]

1. Elastic potential energy (D. EEl)

In this situation, the spring is compressed with the toy on top of it. The toy is stationary, so it does not have kinetic energy. However, the spring is compressed, so it does have elastic potential energy, given by:

$E_{EL}=\frac{1}{2}kx^2$

where k is the spring constant and x is the compression of the spring.

2. Gravitational potential energy (C. Eg)

In this situation, the spring has been released, so it returns to its natural position, so its elastic potential energy is zero. The toy is also stationary, since it is at its top position, where its velocity is zero, so its kinetic energy is also zero. However, the toy is now at a certain height h above the spring, so it has gravitational potential energy given by:

$E_g = mgh$

where m is the mass of the toy and g is the gravitational acceleration.

3. Gravitational potential and kinetic energy (A. Eg and EK)

In this situation, the toy is falling: so, it is moving with a certain speed v, so it has kinetic energy given by

$E_k = \frac{1}{2}mv^2$

Also, since it is at a certain height above the spring, it still has some gravitational potential energy, as in the previous point.

4. Gravitational potential energy (C. Eg)

The jumper is standing on the bridge, so it has gravitational potential energy given by its height h above the ground:

$E_g=mgh$

where m is the mass of the jumper.

5. This exercise has the same text of the previous one.

8 0

3 years ago

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

4 years ago