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Brut [27]
3 years ago
10

What can accurately be said about a resultant wave that displays both reinforcement and interference? A. Maximum interference oc

curs where the troughs of the two cmponent waves are in phase B. The crests of the two compnent waves are in phase C. The component waves have different frequencies D. Molecules remain in their normal positions at spots reinforcement
Physics
2 answers:
vitfil [10]3 years ago
5 0

The situation that can accurately be said about a resultant wave that displays both reinforcement and interference is that crests of the two component waves are in phase. The answer is letter B.

ziro4ka [17]3 years ago
5 0

What can accurately be said about a resultant wave that displays both reinforcement and interference?

A. The crests of the two component waves are in phase where interference occurs in the resultant wave.

B. The component waves have different frequencies.

C. Molecules remain in their normal positions at spots of reinforcement.

D. Maximum interference occurs where the troughs of the two component waves are in phase.

THE CORRECT ANSWER FOR THIS QUESTION IS B GOT 100% ON THE TEST

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Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

V=\frac {1}{3}\times \pi\times r^2\times h

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Also differentiating the expression of volume w.r.t. time as:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given: \frac {dV}{dt} = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

3 0
2 years ago
What is Marie's instantaneous speed at 20 minutes in miles/min?
AVprozaik [17]

Answer:

0.25miles/min

Explanation:

Instantaneous speed of a person or an object is its speed at a particular moment usually at a period of time.

The speedometer of a car reports the instantaneous speed.

 It can be mathematically expressed as;

        Instantaneous speed  = \frac{distance}{time}

At 20min the distance covered is 5miles;

    Instantaneous speed  = \frac{5 miles }{20mins}   = 0.25miles/min

8 0
3 years ago
How do you calculate energy lost due to friction in an experiment?
Snowcat [4.5K]

Answer:

A treadmill get it? but its   Ff * d cos theta

Explanation:

6 0
3 years ago
Please help! Calculate velocity. Show all work!
Eduardwww [97]

Answer:

v = 23.66 m/s

Explanation:

recall that one of the equations of motion may be expressed:

v² = u² + 2as,

Where

v = final velocity (we are asked to find this)

u = initial velocity = 0 m/s since we are told that it starts from rest

a = acceleration = 0.56m/s²

s = distance traveled = given as 500m

Simply substitute the known values into the equation:

v² = u² + 2as

v² = 0 + 2(0.56)(500)

v² = 560

v = √560

v = 23.66 m/s

4 0
3 years ago
Why is the following situation impossible? A skater glides along a circular path. She defines a certain point on the circle as h
Arturiano [62]

Answer:

A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin.

So here in circular motion of the skater we can see that the total path length of the skater is along the arc of the circle while we can say that displacement is defined as the shortest distance between initial and final position of the object.

So it is not possible in any circle that arc-length is less than the chord joining the two points on the circle

As we know that arc length is given as

L = R\theta

length of chord is given as

L_c = 2Rsin(\frac{\theta}{2})

so here

L > L_c

R\theta > 2R sin(\frac{\theta}{2})

so we have

\frac{\theta}{2} > sin(\frac{\theta}{2})

6 0
2 years ago
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