Answer:
At low pressure-
At high pressure-
Explanation:
Initial speed,
Final speed,
Net horizontal force due to rolling friction
mg where m is mass, g is acceleration due to gravity,
is coefficient of rolling friction
From kinematic relation,
For each tire,
Making
the subject
Under low pressure of 40 Psi, d=18 m
Therefore,
At a pressure of 105 Psi, d=93.7
Therefore,
Answer:
a.After
second Mr Comer's speed

b.Distance travelled by Mr.Comer in
seconds

Explanation:
a. Lets recall our first equation of motion 
Now we know that
,
and

Plugging the values we have.




Then Mr.Comer's speed after
sec

b.
Lets find the distance and recall our third equation of motion.

So
distance covered.
Dividing both sides with 2a we have.

Plugging the values.


So Mr.Comer will travel a distance of
.
When solar radiation reaches the Earth it quickly dissipates as most of the radiation and UV rays are blocked by ozone layer, but more radiation and UV rays are able to get through because of global warming.
Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law




Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula


Hence, The speed of space station floor is 49.49 m/s.
We know, the ideal gas equation,
P1V1 / T1 = P2V2 / T2
Here, P1 = 760 mm
V1 = 10 m3
T1 = 27 + 273 = 300 K
P2 = 400 mm Hg
T2 = -23 + 273 = 250 K
Substitute their values,
760*10 / 300 = 400 * V2 / 250
25.33 * 250 = 400 * V2
V2 = 6333.333/ 400
V2 = 15.83
In short, Your Answer would be approx. 15.83 m3
Hope this helps!