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2 years ago

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A periodic wave travels from one medium to another. Which pair of variables are likely to change in the process? A. velocity and

wavelength B. velocity and frequency C. frequency and wavelength D. frequency and phase E. wavelength and phase

1 answer:

dezoksy [38]2 years ago

5 0

Answer:

A. velocity and wavelength

Explanation:

When a wave travels from one medium to another it undergoes a change in direction and this is referred to as refraction.
Refraction is the bending of a wave or a change in direction of a wave as it travels from one medium to another. Refraction is accompanied by change in velocity and wavelength of a wave.
Velocity of a wave is proportional to the wavelength of the wave therefore, if the velocity of a wave changes the the wavelength will also change proportionally to the velocity.

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The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle e

olga55 [171]

Initial speed of the particle is

$1.497\sqrt{(1/m)/(kq^{2} /d)}$

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2 years ago

At the instant that the velocity of the crate is v⃗ =(3.40m/s)ι^+(2.20m/s)j^, what is the instantaneous power supplied by this f

Zigmanuir [339]

I found some missing information about this problem online. We are given the force:
$F = F =(-7.50N)i +(3.00N)j$
Power is defined as the rate of doing work.
This is the formula:
$P= \frac{dW}{dt}$
Where P is power, W is work.
Work is defined as:
$W=F\cdot r$
F is the force and r is the displacement.
If we assume that force is not changing (it's constant) with time we get:
$P= \frac{dW}{dt}=F \frac{dr}{dt}=F\cdot v$
Keep in mind that both force and velocity are vectors, so we have to multiply each component separately.
Finally, we get:
$P=F_i\cdot v_i+F_j\cdot v_j=(-7.50N)(3.40\frac{m}{s})+(3.00N)(2.20\frac{m}{s})=
-18.9 W$

5 0

2 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Vladimir79 [104]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

5 0

2 years ago

"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

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2 years ago

Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (

Pepsi [2]

Answer:The higher up an object is the greater its gravitational potential energy. The larger the distance something falls through the greater the amount of GPE the object loses as it falls. As most of this GPE gets changed into kinetic energy, the higher up the object starts from the faster it will be falling when it hits the ground. So a change in gravitational potential energy depends on the height an object moves through.

Explanation: Lifting an apple up 1 metre is easier work than lifting an apple tree the same height. This is because a tree has more mass, so it needs to be given more gravitational potential energy to reach the same height.

6 0

2 years ago