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3 years ago

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Now instead a 3-mw laser of wavelength λ = 760 nm is directed into the interferometer (with the mirror m1 at the displacement ca

lculated in part a). how much power is detected at d1?

1 answer:

slavikrds [6]3 years ago

7 0

power received will always remain same and does not depends on the distance

it will be same as the power of source

$P = 3 mW$

now the energy released by one photon is given as

$E = \frac{hc}{\lambda}$

$E = \frac{6.6* 10^{-34}* 3 * 10^8}{760 * 10^{-9}}$

$E = 2.6 * 10^{-19} J$

now let say N photons per second released

$N * 2.6 * 10^{-19} = 3 * 10^{-3}$

$N = 1.15* 10^{16} per\: second$

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A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the

REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0

3 years ago

Which of the following examples illustrates static friction in action?

lilavasa [31]

I think its d. but im not sure

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3 years ago

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A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th

vladimir1956 [14]

Answer:

Force, $F=2.27\times 10^4\ N$

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

$F.d=\dfrac{1}{2}m(v^2-u^2)$

Finally, it stops, v = 0

$F.d=-\dfrac{1}{2}m(u^2)$

$F=\dfrac{-mu^2}{2d}$

$F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}$

F = -22713.92 N

$F=2.27\times 10^4\ N$

So, the magnitude of the force that stops the bullet is $2.27\times 10^4\ N$

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3 years ago

When a pitcher is throwing a ball, is he applying positive or negative work to the ball?

Romashka [77]

A.Positive as work=force×distance (assuming that distance is always a constant) the work will always be positive as the force is always positive(this is because force=mass×acceleration where the acceleration is always positive unless the mass is being slowed down)

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In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:

fenix001 [56]

Acceleration is given by:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) $-20.5 m/s^2$

Let's convert the quantities into SI units first:

$u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s$

$v=1600 km/h \cdot \frac{1000 m/km}{3600 s/h} =444.4 m/s$

t = 4.0 min = 240 s

So the acceleration is

$a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2$

B) $-3.8 m/s^2$

As before, let's convert the quantities into SI units first:

$u = 444.4 m/s$

$v=321 km/h \cdot \frac{1000 m/km}{3600 s/h} =89.2 m/s$

t = 94 s

So the acceleration is

$a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2$

C) $-53.0 m/s^2$

For this part we have to use a different formula:

$v^2 - u^2 = 2ad$

where we have

v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

a is the acceleration

d = 75 m is the distance covered

Solving for a, we find

$a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2$

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3 years ago