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lianna [129]
3 years ago
15

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-l

ike nature of the board to help propel him into the air. Assume that the diver's motion is essentially vertical. He leaves the board, which is 3.0 m above the water, with a speed of 6.3 m/s. The diver is in the air for 1.7 s from when he left the board to the water. What is the speed of the diver when he reaches the water?
Express your answer with the appropriate units.
Physics
2 answers:
Artyom0805 [142]3 years ago
8 0

Answer:

final speed with which it will hit the water is v = 9.92 m/s

Explanation:

Here we know that diver's motion is essentially in air

So the motion of diver is under gravity at uniform acceleration

so we can use kinematic in order to find the final velocity

so we will have

v_f^2 - v_i^2 = 2ad

v_f^2 - 6.3^2 = 2(9.8)(3)

so we will have

v_f = 9.92 m/s

so final speed with which it will hit the water is v = 9.92 m/s

Ksenya-84 [330]3 years ago
3 0

Answer:

10.4 m/s

Explanation:

The problem can be solved by using the following SUVAT equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the diver in the problem, we have:

u=+6.3 m/s is the initial velocity (positive because it is upward)

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

By substituting t = 1.7 s, we find the velocity when the diver reaches the water:

v=+6.3 + (-9.8)(1.7)=-10.4 m/s

And the negative sign means that the direction is downward: so, the speed is 10.4 m/s.

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Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
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Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

4 0
3 years ago
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