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3 years ago

8

"In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the scr

een-to-slit distance D must be changed to"

1 answer:

max2010maxim [7]3 years ago

7 0

Answer:

To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to a value 2D that is twotimes D

You might be interested in

What is the kinetic energy of a 1 kg pie if it is thrown at 10 m/s?

nignag [31]

Given

mass (m) = 1 kg

velocity (v) = 10 m/s

kinetic energy ( ke) = ?

,we know

K.E =1/2 m v²

= 1/2 * 1 * 10²

= 100/2

=50 joule

hope it helps :)

7 0

3 years ago

A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters

emmasim [6.3K]

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

Explanation:

For a charged particle, the force on the particle is given as

$F = \frac{mv^{2} }{r}$

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = $2\pi r$

Work on a particle = 2πrF = $2\pi mv^{2}$

This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.

3 0

3 years ago

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

<u></u>

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.

crimeas [40]

Answer:

Explanation:

Given:

length of ladder $r_L = 14m$

weight of ladder $F_L = 490N$

position of firefighter $r_F = 3.8m$

weight of firefighter $F_F = 820N$

angle of ladder $\alpha = 63$

Unknown:

force of the wall on the ladder $F_W$

force of friction on base of ladder $F_R$

normal force on base of ladder $F_N$

From the free body diagram of the sketch you get 3 equations:

$F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0$

Solving the equations gives:

$F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}$

a)

$F_R = 238N\\ F_N = 1310N$

b)

$F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3$

c) Using the result from b and solving for $r_F$

$\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m$

4 0

3 years ago

Why does gravity hold objects on earth?

LUCKY_DIMON [66]

Gravity is an attractive force that works to pull objects together. If 2 objects are close the gravitational pull will be stronger
Mass and distance determine gravity. The farther two things are away from each other, the weaker the gravitational forces are, the less mass an object has the less gravitational force it exerts

6 0

3 years ago

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