Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
<span>Answer:
The temperature doesn't affect the evaporation rate, but affects on how much of water a parcel of air can contain when saturated which is known by the absolute humidity. Hurricanes are usually happening when the temperature of the sea water west of the Cape Verde islands is over 27 degrees Celsius. If ahead of the path of a hurricane, the sea water temperature drops then it will be less moisture in the air and perhaps the hurricane will fade out. But it is not as simple. How strong a tropical storm is is relative to the difference of temperture between ground level and the top of the troposphere. The greater the difference, the faster the air will rise and the deeper the pressure will be, forcing surrounding air to rush in, thus forming a hurricane force wind. Then there is the fact that the wet adiabatic lapse rate is about half that of dry air. It means that rising moist air cools down slower and therefore rises higher. Hence water is the true fuel of bad weather. But it can't be isolated from the fact that the difference of temperature must be great too. What we often forget is that the tropopause (the border to the stratosphere) is much higher over the equator and therefore, much colder than e.g. the poles.</span>
Given :
Walk in forward direction is 30 m .
Walk in backward direction is 25 m .
To Find :
The distance and displacement .
Solution :
We know , distance is total distance covered and displacement is distance between final and initial position .
So , distance travelled is :
D = 30 + 25 m = 55 m .
Now , we first move 30 m in forward direction and then 25 m in backward direction .
So , displacement is :
D = 30 - 25 m = 5 m .
Therefore , distance and displacement covered is 55 m and 5 m respectively .
Hence , this is the required solution .
Answer:
532 millimeters of mercury
Explanation:
In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:
1 atm = 760 mm Hg
Therefore, we can solve the problem by setting up the following proportion:
Solving for x, we find
Answer:
3a, 2b,4c,1d
Explanation:
what do I need to explain just something you know
