The common #2 pencil is 7 1/2 long with a wooden shaft measuring about 6 3/4
The water pressure at this depth and the total pressure due to water and atmosphere are 10.3 x 10⁵ Pa and 11.31× 10⁵ Pa.
<h3>What is pressure?</h3>
The pressure is the amount of force applied per unit area.
Atmospheric pressure, Patm =1.01×10⁵ Pa
Density of water, ρ=1030 kg/m³
Depth, h=100 m
Pressure =ρgh
P = 1030×10×100
P = 10.3 x 10⁵ Pa.
Total pressure, P=Po +ρgh
P=1.01×10⁵ + 1030×10×100
P=11.31× 10⁵ Pa
Hence, total pressure is 11.31× 10⁵ Pa.
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The most useful meteorological measurement for forecasting freezing precipitation is b. radiosonde soundings
<h3>
Radiosonde </h3>
At high altitudes, radiosondes are battery-powered telemetry sensor bundles that detect altitude, pressure, temperature, relative humidity, wind (both speed and direction), and cosmic ray measurements. They are commonly taken into the atmosphere by weather balloons.
Rawindsonde is an acronym for radar wind sonde, a type of radiosonde that tracks its position as it rises through the sky to provide wind speed and direction. Another type of radiosonde is one that falls to the ground after being released from an aircraft, as opposed to being carried by weather balloons. The term "dropsondes" is used to describe this group of radiosondes. The majority of operational atmospheric data assimilation methods depend heavily on radiosondes.
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Answer is 6 tires.
This is a projectile question.
First make sure units are consistent - express speed in m/s.
20 km/h = 20000m / 3600 s = 5.56 m/s
Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.
Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:
Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s
An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:
Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s
Total trip time:
0.19 x 2 = 0.38s
Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:
v = d / t
5.22 = d / 0.38
d = 1.98m
Now divide this total distance by the length of an individual tire to find the number of tires he will clear:
1.98 / 0.3 = 6.6 tires
Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).
Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
