<u>Answer:</u>
<h3>During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.</h3>
<u>Sources:</u>
-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads
and
-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262
I hope this helps you! ^^
Answer:
The velocity is 
Explanation:
From the question we are told that
The first distance is 
The first speed is 
The second distance is 
The second speed is 
Generally the time taken for first distance is
The time taken for second distance is
The total time is mathematically represented as
=> 
=> 
Generally the constant velocity that would let her finish at the same time is mathematically represented as
=> 
=>
Explanation:
A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path subjected to gravity force only.
Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.
In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.
The answer is B(t) = constants x I(t)
Please take precaution on the point that it is an independent field of its radial position, if the point is measured well in the solenoid. (also the radial position is the axis of its symmetry)
Answer: a) 456.66 s ; b) 564.3 m
Explanation: The time spend to cover any distance a constant velocity is given by:
v= distance/time so t=distance/v
The slower student time is: t=780m/0.9 m/s= 866.66 s
For the faster students t=780 m/1,9 m/s= 410.52 s
Therefore the time difference is 866.66-410.52= 456.14 s
In order to calculate the distance that faster student should walk
to arrive 5,5 m before that slower student, we consider the follow expressions:
distance =vslower*time1
distance= vfaster*time 2
The time difference is 5.5 m that is equal to 330 s
replacing in the above expression we have
time 1= 627 s
time2 = 297 s
The distance traveled is 564,3 m
