HEY CAN ANYONE ANSWER DIS QUESTION PLS!!!!!!!!

A medical treatment where the individual breathes in pure oxygen while in a pressurized room or through a tube. This increased p

Leto [7]

Hyperbaric Oxygen Therapy (HBOT) is a medical treatment where the individual breathes in pure oxygen while in a pressurized room or through a tube. The HBOT treatment includes breathing 100% oxygen while under increased atmospheric pressure. It is used for decompression sickness by scuba divers.

5 0

3 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

so

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago

What is the difference between brake horsepower and thrust horsepower?

arlik [135]

1. HP is the output horsepower rating of an engine, while Brake horse power is the input brake horsepower of an engine. ... Brake horse power is the measurement of an engine's power without any power losses, while HP is less the power losses Brake horse power

5 0

3 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

Which sediment size would allow water to flow through at the fastest rate?

xeze [42]

The sediment size that would allow water to flow through at the fastest rate are pebbles. Sediment is a naturally occurring material that is broken down by processes of weathering and erosion, and is consequently transported by the action of wind, or ice, and or by the force of gravity acting on the particles. Pebble is a clast of rock with a particle size of 2 to 64 millimeters based on the scale of sedimentology.

4 0

2 years ago