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3 years ago

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When a high-power laser is used in the Earth's atmosphere, the electric field associated with the laser beam can ionize the air,

turning it into a conducting plasma that reflects the laser light. In dry air at 0°C and 1 atm, electric breakdown occurs for fields with amplitudes above about 3.00 MV/m.
(a) What laser beam intensity will produce such a field?
W/m2

(b) At this maximum intensity, what power can be delivered in a cylindrical beam of diameter 3.44 mm?
W

1 answer:

Marianna [84]3 years ago

4 0

Find the below attachment

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a 5.5 g dart is fired into a block of wood with a mass of 22.6 g. the wood block is initially at rest on a 1.5 m tall post. afte

IgorLugansk [536]

<span>From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block.
m1 = 5.5g

$5.5g$ x $\frac{1kg}{1000g}= 0.0055kg$

m2 = 22.6g

$22.6g$ x $\frac{1kg}{1000g}= 0.0226kg$

So now we settled that we can set our given as:
M1 = .0055 kg
v1i = ?
M2 = 0.0226 kg
v2i = 0 m/s
dx = 2.5 m
dy = -1.5 m

Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf.

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x.
</span>

dy = -1.5 m

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

<span>We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity.

$dy = v_{iy}t + \frac{1}{2} at^{2}$ </span>

$1.5 = (0)t+\frac{1}{2} (9.8)t^{2}$

<span> $1.5 = \frac{1}{2} (9.8)t^{2}$

$\frac{(2)(1.5)}{9.8}=t^{2}$

$\frac{(3)}{9.8}=t^{2}$

$\sqrt{0.3061} = \sqrt{t^{2}$

$0.553s = t$

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
$v_{x} =d_{x}t$ and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
$v_{x} =d_{x}t$
$v_{x} =(2.5m)(0.553s)$
$v_{x} =4.52m/s$

Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

From this we can derive the equation for v1i by isolating it.

$v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}$

Now let's put in all our given and what we solved:

$v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}$

$v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}$

$v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}$

$v_{1i}= 23.09m/s$

The initial speed of the dart is 23.09 m/s or 23.10 m/s.</span>

7 0

4 years ago

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

$F = \sqrt{F_{x}^{2} +F_{y}^{2} }$

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

$15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]$

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3 years ago

Can I have some tips, tricks, and ideas for dropping an egg 12 feet in the air onto a hard surface without it breaking. It HAS t

lara [203]

Have it land straight up

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The moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is

sleet_krkn [62]

Answer:

I = I₀ + M(L/2)²

Explanation:

Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.

The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.

The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀

The distance between the two axes is L/2 (total length of the rod divided by 2

From the parallel axis theorem we have

I = I₀ + M(L/2)²

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3 years ago

In the 1960s what dance was being created in the west coast

DerKrebs [107]

The dance that was being created was Hip Hop dance!

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