Answer:
Height reached will be 28.35 m
Explanation:
Here we can use the work energy theorem to find the maximum height
As we know by work energy theorem
Work done by gravity + work done by friction = change in kinetic energy
![-mgh - F_f h = 0 - \frac{1}{2}mv_i^2](https://tex.z-dn.net/?f=-mgh%20-%20F_f%20h%20%3D%200%20-%20%5Cfrac%7B1%7D%7B2%7Dmv_i%5E2)
now we will have
![-1.60(9.8)(h) - 0.900(h) = - 470](https://tex.z-dn.net/?f=-1.60%289.8%29%28h%29%20-%200.900%28h%29%20%3D%20-%20470)
![-16.58 h = -470](https://tex.z-dn.net/?f=-16.58%20h%20%3D%20-470)
![h = 28.35 m](https://tex.z-dn.net/?f=h%20%3D%2028.35%20m)
so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction
Answer:
Your question was incomplete so here is the complete question and answer.
Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)
a) plain water
b) 5-7 percent glucose solution
c) Glucose polymer solution of 6-8 percent
d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.
Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.
Explanation:
Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.
Cody ...
Everything on this page is solved with the SAME formula !
Distance = (speed) x (time) .
Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.
'Velocity' is speed AND THE DIRECTION OF THE MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.
Your teacher may not be happy if you talk about this on your homework,
but that's too bad. Just don't say "velocity" in any of your answers.
Say "speed", and if the teacher complains about that, then it's time to
let the teacher have it with both barrels.
1). Speed = (distance covered) / (time to cover the distance)
2). Speed = (distance covered) / (time to cover the distance)
3). Distance = (average speed of travel) x (time traveling at that speed)
4). Time to cover the distance = (distance) / (speed)
5). Car's speed = (distance the car covered) / (time the car took)
Sprinter speed = (distance the sprinter covered) / (time the sprinter took)
Calculate the car's speed.
Calculate the sprinter's speed.
... Look at the two speeds.
Decide which one is faster.
... Subtract the slower one from the faster one.
The difference is the answer to "by how much?" .
6). Distance = (speed) x (time spent moving at that speed)
7). Average speed = (TOTAL distance covered)
divided by
(time to cover the TOTAL distance).
Answer:
B'= 3.333 B
Explanation:
Lets take
Initial area = A
Magnetic field = B
The area after compression
A'=0.3 A
Magnetic field = B'
We know that flux ,Ф
Ф = B A
Given that flux is constant so
B A = B' A'
B A=B' x 0.3 A
B'= 3.333 B
It means that magnetic field will increase.
(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.
R = sqrt ((20 knots)² + (17 knots)²)
R = sqrt (400 + 289)
R = 26.24 knots
The direction of the wind will have to be angle between the boat and the resultant.
cos θ = (20 knots)/(26.24 knots)
θ = 40.36°
Hence, the direction is 40.36° east of north.
(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero.