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4 years ago

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Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum

destructive interference is to occur at point P, the two beams must travel paths that differ by

1 answer:

Leto [7]4 years ago

6 0

Answer:

the two beams must travel paths that differ by one-half of a wavelength.

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A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.

The answer is false

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3 years ago

How many type of linear expansion physic

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three types

The three types of thermal expansions are Linear expansion , Superficial expansion and Cubical expansion.

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3 years ago

If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate

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0.5 x 9.81 x 3^2 = 44m

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3 years ago

Read 2 more answers

Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th

lozanna [386]

Answer:

Part a)

Velocity of sled

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

Velocity of sled

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

$0 = m_1(v - s) + (m_2 + M)v$

$v = \frac{m_1 s}{m_1 + m_2 + M}$

so velocity of the sled + other person is

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s$

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f$

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f$

Now we have

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s$

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

here we know all values of speed as we found it in part a) and part b)

4 0

3 years ago

a bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s² for 10s. Find the distance moved by it during this interval o

BaLLatris [955]

Answer:

90 m

Explanation:

Acceleration, $a=\frac {v-u}{t}$ where v and u are final and initial velocities respectively, t is the time taken

Substituting $1 m/s^{2}$ for a, 4 m/s for u and 10 s for t then

1*10=v-4

v=14 m/s

From kinematic equations

$v^{2}=u^{2}+2as$

Making s the subject then

$s=\frac {v^{2}-u^{2}}{2a}=\frac {14^{2}-4^{2}}{2\times 1}=90 m$

6 0

3 years ago