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insens350 [35]
3 years ago
14

Two waves with equal amplitude meet each other, resulting in a wave zero amplitude. Which phenomenon explains this result?

Physics
2 answers:
jeka57 [31]3 years ago
4 0

Answer:

(D) Interference

Explanation:

Interference is a phenomenon characteristic for waves. When two waves meet, they interfere and result in a new wave. The precise shape of the resultant wave depends on the amplitude, frequency and phase of the constituent waves. In a most basic setting, two types of interference occur: constructive (resultant amplitude increases) and destructive (resultant amplitude decreases).

Lynna [10]3 years ago
4 0
You answer to this question is D
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256N. I love physics so.... yerp
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Directions: I answer all the questions listed below.
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Answer:

The compression ratio is 10

Explanation:

Given

V_d = 450cm^3 --- swept volume

V_c = 50cm^3 --- compression volume

Required

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CR = \frac{V_d + V_c}{V_c}

CR = \frac{450+50}{50}

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A ball thrown with 50N of force accelerates at 25 m/s2.  What is the mass of the ball?
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F=ma
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7 0
4 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
A car traveling at 31 m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is its acceleration
N76 [4]
Vf=V1+at
0=31+a(7)
-31/7=a
a=-4.43m/s^2

Hope this helps.
8 0
3 years ago
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