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3 years ago

Which part of the wave is changed when there is interference?

Physics

2 answers:

slava [35]3 years ago

8 0

THE AMPLITUDE. (this is the first option on idgenuity)

Rasek [7]3 years ago

3 0

When 2 waves interefere (or collide with eachother), it usually affects the crest of the wave. If both waves collide with both crests, it will create an amplified crest, and the waves will pass through eachother afterwards. If a trough of a wave meets a crest, it will cause the crest to be lowered shortly before both continue on.

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A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho

dlinn [17]

Answer:

$x =\frac{-R}{2}$

Explanation:

From the question we are told that mass

Thin layer radius $= 2R$

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state

Considering the center of mass of both bodies

$xcm=\frac{m*x+m*x)}{2m} =x$

$x =\frac{-R}{2}$

Therefore the enclosing layer moves $x =\frac{-R}{2}$

7 0

3 years ago

A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01 ???? 10

SIZIF [17.4K]

Answer:

the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019

Explanation:

given information

h = 0.2 m

$P_{0}$ = 1.01 x $10^{5}$ Pa

$P_{1} V_{1} = P_{2} V_{2}$

$\frac{V_{2} }{V_{1}} = \frac{P_{1} }{P_{2}}$

$P_{1}$ = $P_{0}$ + ρgh, ρ = 1000 kg/ $m^{3}$

$P_{1}$ = 1.01 x $10^{5}$ Pa + (1000 x 9.8 x 0.2) = 1,0296 x $10^{5}$ Pa

$P_{2}$ = $P_{0}$ = $10^{5}$ Pa

thus,

$\frac{V_{2} }{V_{1}} = 1,0296 x [tex]10^{5}$ / $10^{5}$ = 1.019

8 0

3 years ago

How is the medium of a wave and transmit of a wave similar and different?

zepelin [54]

Answer:

Sound and light are similar in that both are forms of energy that travel in waves. They both have properties of wavelength, freqency and amplitude. Here are some differences: Sound can only travel through a medium (substance) while light can travel through empty space.

3 0

3 years ago

Given a 10-V power supply, would a 20-ohm resistor and a 5-ohm resistor need to be arranged in parallel or in series to generate

Bogdan [553]

Answer:

The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

Explanation:

We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.

We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.

Using ohm's law we have

V = I*R

V= voltage

I= current

R= resistance

10 = 2.5*R

R = 10/2.5 = 4ohm

that means we need a total of 4ohm resistance from these two resistors.

since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.

$\frac{1}{R} = \frac{1}{R1} +\frac{1}{R2}\\ = \frac{1}{20} +\frac{1}{5}\\ \\ = \frac{1+4}{20} =\frac{1}{4}$

R(net) =4ohm

Now the orientation of the resistors are in parallel so the current will be divided.

we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.

We know that the voltage in parallel remains same

In 20 ohm resistance

again using ohms law

V = i1*R1

10 = i1*20

i1 = 0.5A

in 5ohm resistor

V=i2*R2

10 = I2*5

i2 =2A

and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.

Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

6 0

3 years ago

A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2

lys-0071 [83]

Answer: sheet of charge

Explanation:

a )

Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.

b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.

C ) Electric field which is almost uniform near the sheet of charge is equal t the following

E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²

E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²

= .3389 x 10³

= 338.9 V / m

spacing between 1 V

= 1 / 338.9 m

= 2.95 X 10⁻3 m

= 2.95 mm.

3 0

3 years ago