A kangaroo jumps up with an initial velocity of

Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes

scoundrel [369]

Answer:

The appropriate solution is "0.597 mm".

Explanation:

Given that:

Wavelength,

$\lambda = 519 \ nm$

or,

$=519\times 10^{-9}$

Distance,

$D=4.6 \ m$

Separated by,

$\beta = 4.0 \ mm$

or,

$=4.0\times 10^{-3}$

As we know,

⇒ $\beta=\frac{\lambda D}{d}$

or,

The separation of two slits will be:

⇒ $d=\frac{\lambda D}{\beta}$

By putting the values, we get

$=\frac{(519\times 10^{-9})(4.6)}{4.0\times 10^{-3}}$

$=5.97\times 10^{-4}$

or,

$=0.597 \ mm$

8 0

3 years ago

The Earth is instantly replaced in its orbit by a speck of dust. Which statement best describes the subsequent orbital motion of

Alik [6]

The correct statement is:

The dust particle will move to a larger orbit and orbit the Sun in more than 1 year.

In fact, the dust particle has smaller mass than the Earth, therefore the gravitational attraction exerted by the Sun on the dust will be smaller, according to the formula of the gravitational force:

$F=G\frac{M_S m}{r^2}$

where G is the gravitational constant, MS is the Sun mass, m is the mass of the Earth (at first) and then of the dust particle (later), and r is the distance from the Sun. When we replace the Earth with the dust particle we see that m decreases, so the gravitational force F decreases, and the particle will move to a larger orbit. Then its orbital period will increase, according to the third law of Kepler, which states that the square of the orbital period is proportional to the cube of the orbital radius:

$T^2 \sim r^3$

therefore, since the orbit is larger, the orbital period is greater.

5 0

3 years ago

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

MAVERICK [17]

Answer:

$U_{b}=+7.3*10^{-8}J$

Explanation:

When a particle moves from a point where the potential energy is Ua to point where it is Ub,the change in potential energy is is equal to work done

So

$W_{a-b}=U_{a}-U_{b}\\ U_{b}=U_{a}-W_{a-b}\\$

Where Wa-b here is negative this means Ub is greater Ua. Therefore potential energy increases

So

$U_{b}=(+5.4*10^{-8}J )-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J$

7 0

3 years ago

when the total amount of a quantity is the same before and after some event occurs, we say that the quantity is (blank)

Effectus [21]

Ah hah ! That description fits things like mass, linear momentum,
energy, angular momentum, electric charge, and politicians. They
never appear or disappear, they just move from place to place.
We say that those quantities are conserved.

8 0

3 years ago

How do you identify the zero reference position for calculating mechanical potential energy? What is the most common zero refere

ANTONII [103]

<span>The zero reference position should be the position the system has the least potential energy. As an object falls, its potential energy is converted into kinetic energy. Gravity's ability to do work (potential energy) reaches zero when it hits the ground and stops falling, hence the ground is the most common zero reference position.</span>

8 0

3 years ago