Answer:you in connections too?
Explanation:
Answer:
a. a=33.34ms⁻², V=164.4m/s
Explanation:
Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.
Below are the data given
Distance, s=404.5m,
time taken,t=4.922secs
Using the equation
S=ut+1/2at²
where u is the initial velocity and u=0
Making the acceleration the subject of the formula, we arrive at
a=2s/t²
a=(2*404.5)/4.922²
a=33.34ms⁻².
To determine the velocity, we use
V=u+at
V=0+33.34ms⁻² *4.922sec
V=164.4m/s
It totally depends on what kind of wave you're talking about.
-- a sound wave from a trumpet or clarinet playing a concert-A pitch is about 78 centimeters long ... about 2 and 1/2 feet. This is bigger than atoms.
-- a radio wave from an AM station broadcasting on 550 KHz, at the bottom of your radio dial, is about 166 feet long ... maybe comparable to the height of a 10-to-15-story building. This is bigger than atoms.
-- a radio wave heating the leftover meatloaf inside your "microwave" oven is about 4.8 inches long ... maybe comparable to the length of your middle finger. this is bigger than atoms.
-- a deep rich cherry red light wave ... the longest one your eye can see ... is around 750 nanometers long. About 34,000 of them all lined up will cover an inch. These are pretty small, but still bigger than atoms.
-- the shortest wave that would be called an "X-ray" is 0.01 nanometer long. You'd have to line up 2.5 billion of <u>those</u> babies to cover an inch. Hold on to these for a second ... there's one more kind of wave to mention.
-- This brings us to "gamma rays" ... our name for the shortest of all electromagnetic waves. To be a gamma ray, it has to be shorter than 0.01 nanometer.
Talking very very very very roughly, atoms range in size from about 0.025 nanometers to about 0.26 nanometers.
The short end of the X-rays, and on down through the gamma rays, are in this neighborhood.
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and 
The sum is represented as
For the the values given to us the sum is calculated as
Now the since the uncertainity inthe sum is 
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals 
meters
Answer : 63 km/h
Explanation :
Outgoing distance = 300 km.
Outgoing speed = 93 km/h.
Break time = 1 h
Return distance = 300 km
Return speed = 56 km/h
Outgoing time = outgoing distance / outgoing speed
Outgoing time = 300 km / 93 km/h = 3.225806451612 h
Return time = return distance / return speed
Return time = 300 km / 56 km/h = 5.357142867142 h
Total distance =. Outgoing distance + return distance travelled
Total distance = 300 km + 300 km = 600 km
Total time = outgoing time + break time + return time
Total time = 3.225806451612 h + 1 h+ 5. 357142867142 h = 9.582949308754 h
Average speed = total distance / total time
Average speed = 600 km / 9.582949308754 h
Average speed = 62.61120462042 km/h = 63 km/h
