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vovikov84 [41]
3 years ago
5

If two light waves are coherent a) their amplitudes are the same their phase difference is constant their frequencies are the sa

me b) c) d) their wavelengths are the san the difference in their frequencies is constant
Physics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

a) their amplitudes are the same their phase difference is constant their frequencies are the same

Explanation:

Coherent waves are the waves that have constant phase difference, equal frequency, amplitude and waveform.

Frequency denotes the number of cycles a wave completes in one second.

Amplitude is the maximum height that the wave reaches.

Waveform is the two dimensional representation of a wave in graphical form.

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Atom X contains seven protons and seven neutrons. Atom Z contains seven protons and eight neutrons. Which of the following state
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It is A. They ate isotopes
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A clinical psychologist is one type of pure basic research psychologist . True are false
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4 years ago
A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

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3 years ago
Which of the following occurs when the parasympathetic nervous system PNS is activated
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5 0
4 years ago
A 4.2-g bullet is fired at a speed of 370 m/s into a stationary lead block, which moves a distance of 1.2 m before coming to res
Natali5045456 [20]

Answer:

(a) F = 239.575 N (b) t = 0.00649s or 6.49 ms

Explanation:

(a) By law of energy conservation, the bullet kinetic energy will be transferred to work done on stopping it from moving.

Formula for Kinetic Energy E_k = \frac{mv^2}{2} where m is bullet mass, v is the velocity

Formula for work W = FS where F is the average force and S is the distance travelled.

E_k = W

\frac{mv^2}{2} = FS

F = \frac{mv^2}{2S}

Substitute m = 4.2 g = 0.0042 kg, v = 370 m/s and S = 1.2 (m)

F = \frac{0.0042*370^2}{2*1.2} = 239.575 N

(b) If the force is constant, since the mass is constant and F = ma according to Newton's 2nd law, the acceleration on bullet is also constant

a = \frac{F}{m} = \frac{239.575}{0.0042} = 57041.67 (m/s^2)

We also have v(t) = v_0 + at

At the time the bullet is coming to rest, v(t) = 0, a = -57041.67 m/s^2

Therefore, 0 = 370 - 57041.67t

t = \frac{370}{57041.67} = 0.00649 s = 6.49 ms

3 0
4 years ago
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