Coulomb's law:

Force = (8.99×10⁹ N m² / C²) · (charge₁) · (charge₂) / distance²

= (8.99×10⁹ N m² / C²) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C) / (1.0 m)²

= (8.99×10⁹ x 1×10⁻¹² / 1.0) N

= 8.99×10⁻³ N

= 0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses'; don't need it.
 Mass has no effect on the electric force between them.

-- 'frictionless table'; don't need it.
 Friction has no effect on the force between them,
only on how they move in response to the force.

7 0

3 years ago

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

3 0

3 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

3 years ago