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4 years ago

Which of the following forms an interference pattern when directed toward two suitably-spaced slits?

Physics

1 answer:

alekssr [168]4 years ago

6 0

Answer:

Option d

Explanation:

Interference is the phenomenon in which two or more waves having the same frequency combines or cancel out each other to form a resulting wave with the amplitude equals the sum of the amplitudes of the waves that are combined.

Thus it is clear that the particle with the wave nature can produce interference pattern. Hence, electron, light, sound, all of these produces interference pattern when directed towards the slits that are suitably spaced.

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1 point

docker41 [41]

Answer:

what do the choices look like

Explanation:

3 0

3 years ago

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

Vedmedyk [2.9K]

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

5 0

4 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Gnesinka [82]

Answer:

a). $a_p=-2.39x10^{-12} rad/s^2$

b). $t=1016298.8 years$

c). $T_i=80.58x10^{-3}s$

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

$a_p=\frac{dw}{dt}$

$w=\frac{2\pi}{t}$

$a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}$

$dT=0.0808s$

$dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s$

Replacing

$a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2$

b).

If the pulsar will continue to decelerate at this rate, it will stop rotating at time:

$t=\frac{w}{a_p}$

$w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s$

$t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s$

$t=1016298.8 years$

c).

582 years ago to 2019

1437

$T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s$

5 0

4 years ago

A simple, direct-current electrical circuit is shown below.

kap26 [50]

The direction of current flow is given by an arrow in the electrical circuit.

Several categories are used to categorise electrical circuits. One-way current only flows through a direct-current circuit. As with most residential circuits, an alternating-current circuit is a channel for transmitting electrical current that pulses back and forth numerous times each second.

A battery or generator, a device that provides energy to the charged particles that make up the current, a device that uses current, such as a lamp, an electric motor, or a computer, and the connecting wires or transmission lines make up an electrical circuit.

A source of electrical power, two wires that can transmit current, and a light bulb make up an electrical circuit. Both wires have one end connected to a cell terminal and the other end connected to a light bulb. When the light bulb is turned off, the electrical circuit is disrupted.

To learn more about electrical circuits please visit -
brainly.com/question/1922668
#SPJ1

5 0

2 years ago

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

mafiozo [28]

Answer:

a. $f=1.22*10^{-15} N$

b. $f=53.6*10^{-17} N$

Explanation:

The force existing between two charges is given as

$f=\frac{kq_{1}q_{2}}{r^{2}}$

where q= charge,

k=constant

r= distance between the two charges

Note: this force can either be repulsive or attractive force depending on the charges involve. it is repulsive if they are similar charge and it is attractive if it is opposite charges.

Also the charge of an electron is

$-1.602*10^{-19}$

A. we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{0.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{0.04}\\f_{21}=1.44*10^{-15}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis

for the -5.50nC the distance between them is 0.600m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.6^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.36}\\f_{23}=-(0.22*10^{-15})N i$

this this force will be repulsive force and it points away from the electron i.e points towards the -ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=1.44*10^{-15}-0.22*10^{-15}\\ f=1.22*10^{-15} N$

b. at distance of x=1.20m, this is shown on the diagram below (attachment 2)

we first determine the magnitude force between the -4nC and the electron

$f_{21}=\frac{kq_{1}q_{2}}{r^{2}}\\f_{21}=\frac{9*10^{10} 4*10^{-9} *1.602*10^{-19} }{1.2^{2}}\\f_{21}=\frac{57.67*10^{-18} }{1.44}\\f_{21}=4.0*10^{-17}Ni$

this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

for the -5.50nC the distance between them is 0.4m as can be seen in the diagram the magnitude of the force is

$f_{23} =\frac{kq_{1}q_{2}}{r^{2}}\\f_{23}=\frac{9*10^{10} 5.50*10^{-9} *1.602*10^{-19} }{0.4^{2}}\\f_{23}=\frac{79.3*10^{-18} }{0.16}\\f_{23}=49.6*10^{-17}Ni$

this this force will be repulsive force and it points away from the electron i.e points towards the +ve x-axis.

The total net force on the electron is thus

$f=f_{21}+f_{23}\\ f=4.0*10^{-15}+49.6*10^{-17}\\ f=53.6*10^{-17} N$

8 0

3 years ago