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2 years ago

Cutting down trees can decrease the amount of water vapor entering the air through the processes of ____?

1 answer:

5 0

✿ Transpiration is the process by which water is carried through plants from roots and release water vapor through their leaves.

✿ Cutting down trees can decrease the amount of water vapor entering the air through the processes of Transpiration

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What are two different examples of positive exeleration

sdas [7]

Answer:Accelerating your car to get up to speed on a freeway

An airplane accelerating to take off

Accelerating out of the starting blocks at a track meet

Explanation:

5 0

2 years ago

A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif

Free_Kalibri [48]

Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

Explanation:

First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).

So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

$R=\frac{v^{2}sin(2\theta)}{g}$

Where R is the range of the bullet (this is how far it goes before it has the

same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.

We can solve this equation for theta, so we get:

$gR=v^{2}sin(2\theta)$

$\frac{gR}{v^{2}}=sin(2\theta)$

$sin^{-1}(\frac{gR}{v^{2}})=2\theta$

$\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}$

so now we can substitute the given data:

$\theta=\frac{sin^{-1}(\frac{(9.8m/s^{2})(45.5m)}{(477m/s)^{2}})}{2}$

so we get:

θ=0.05614°

once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

$tan \theta = \frac{h}{45.5m}$

so we can solve this for h, so we get:

$h=45.5m*tan(0.05614^{o})$

which yields:

$h=0.0445m$

$h=4.45cm$

5 0

2 years ago

Fiber-optic cables are used widely for internet wiring, data transmission, and surgeries. When light passes through a fiber-opti

Gwar [14]

After one meter, 3.4% of the light is gone ... either soaked up in the fiber
material or escaped from it. So only (100 - 3.4) = 96.6% of the light
remains, to go on to the next meter.

After the second meter, 96.6% of what entered it emerges from it, and
that's 96.6% of 96.6% of the original signal that entered the beginning
of the fiber.

==> After 2 meters, the intensity has dwindled to (0.966)² of its original level.
It's that exponent of ' 2 ' that corresponds to the number of meters that the light
has traveled through.

==> After 'x' meters of fiber, the remaininglight intensity is (0.966) ^x-power
of its original value.

If you shine 1,500 lumens into the front of the fiber, then after 'x' meters of
cable, you'll have
(1,500) · (0.966)^x
lumens of light remaining.

=========================================

The genius engineers in the fiber design industry would not handle it this way.
When they look up the 'attenuation' of the cable in the fiber manufacturer's
catalog, it would say "15dB per 100 meters".

What does that mean ? Break it down: 15dB in 100 meters is 0.15dB per meter.
Now, watch this:

Up at the top, the problem told us that the loss in 1 meter is 3.4% . We applied
super high mathematics to that and calculated that 96.6% remains, or 0.966.

Look at this ==> 10 log(0.966) = -0.15  <== loss per meter, in dB .

Armed with this information, the engineer ... calculating the loss in 'x' meters of
fiber cable, doesn't have to mess with raising numbers to powers. All he has to
do is say ...

-- 0.15 dB loss per meter

-- 'x' meters of cable

-- 0.15x dB of loss.

If 'x' happens to be, say, 72 meters, then the loss is (72) (0.15) = 10.8 dB .

and 10 ^ (-10.8/10) = 10 ^ -1.08 = 0.083 = 8.3% <== That's how much light
he'll have left after 72 meters, and all he had to do was a simple multiplication.

Sorry. Didn't mean to ramble on. But I do stuff like this every day.

5 0

2 years ago

A man can row a boat at 4kmhr in still water. He rows the boat 2km upstream and 2 km back to his starting place in 2 hours. How

Softa [21]

Answer:

2.83km/hr

Explanation:

Let stream flowing at the rat x km/hr

In still water

Speed of boat=4km/hr

Upstream speed=4-x

Downstream speed=4+x

Total time=2 hr

Time=Distance/speed

According to question

$\frac{2}{4+x}+\frac{2}{4-x}=2$

$\frac{8-2x+8+2x}{(4+x)(4-x)}=2$

$\frac{16}{4^2-x^2}=2$

Using the formula

$(a+b)(a-b)=a^2-b^2$

$4^2-x^2=\frac{16}{2}$

$16-x^2=8$

$16-8=x^2$

$8=x^2$

$x=\sqrt{8}=2.83km/hr$

Hence, the stream is flowing at 2.83km/hr

3 0

2 years ago

When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated

Mashcka [7]

Answer:

It is calculated as Force × perpendicular distance.

Explanation:

Torque is a rotational force and twisting force that can cause an object to rotate in it's axis. This cause angular rotation.

The torque due to gravity on a body about its centre of mass is zero because the centre of mass is the that point of the body at which the force acts by the gravity that is mg.

But if the pivot point of a balance is not at the centre of mass of the balance, it will be FORCE × PERPENDICULAR DISTANCE because at that point, there is no centre in which the force act on a body by gravity. The distance and force will be use to calculate.

6 0

2 years ago