What is the net force acting on this box?

a nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm. how many photons are in the pulse?

Wewaii [24]

A nitrogen laser generates a pulse containing 10.0 mj of energy at a wavelength of 340.0 nm and has 1785 x 10¹⁹ photons in the pulse.

<h3>How many photons are in the pulse?</h3>

Energy of a single photon is

E=hcλ

E=6.626×10⁻³⁴ J s×3×108 m/s /340×10⁻⁹ m

E=6.31×10⁻¹⁹ J

Number of photons in the laser is

n=Total Energy/Energy per photon

n=10⁷×10⁻³J /5.90×10⁻¹⁹J/photon

n= 1785 x 10¹⁹ photons

To learn about photons, refer: brainly.com/question/20912241?referrer=searchResults

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6 0

1 year ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

How much heat is necessary to warm 500g of water from 20°c to 65°c

Vsevolod [243]

Explanation:

so sorry

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8 0

2 years ago

What do you mean by Aquatic habitat? Give 2 examples of Desert habitat

Brilliant_brown [7]

Answer:

hope this answer will help you

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and what do u think about indians

8 0

3 years ago

What is a transverse wave? How do the particles in the medium move in relation to the energy of the wave? Does this wave require

Pani-rosa [81]

A transverse wave is a wave where the particles in the medium move perpendicular (at right angles) to the direction of the source or its propagation (think of a snake slithering through grass) an example of a transverse wave could be a light wave. Light waves for instance don’t need a medium in order to propagate but transverse waves in general do need a medium.

4 0

2 years ago

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