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UkoKoshka [18]
2 years ago
8

6. A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, a

nd her takeoff point is 1.80 m above the pool. (3pt) a) How long are her feet in the air? b) What is her highest point above the board? c) What is her velocity when her feet hit the water?
Physics
1 answer:
mestny [16]2 years ago
5 0

The results of the calculation are;

a) The feet spends 0.41 s in air

b) The highest point above board is 2.62 m

c) The velocity when her feet hit the water is  7.2 m/s

<h3>What is the time spent in air?</h3>

From the data presented;

v = u + at

But v = 0 m/s at the maximum height thus;

0 = 4 - (9.8 * t)

4 = 9.8 * t

t = 4/9.8

t = 0.41 s

b) from;

h = ut - 1/2gt^2

h = (4 * 0.41) - (0.5 * 9.8 * (0.41)^2)

h = 1.64 - 0.82

h = 0.82 m

The total height above board = 0.82 m + 1.8 m = 2.62 m

c) The total time in air is obtained from;

h = ut + 1/2gt^2

u = 0m/s because she dropped off the board

h = 1/2gt^2

2.62 = 0.5  * 9.8 * t^2

t = √2.62/0.5  * 9.8

t = 0.73 seconds

Hence, the velocity when her feet hit the water is obtained from;

v = u + gt

when u = 0 m/s

v = gt

v = 9.8 *  0.73 s

v = 7.2 m/s

Learn more about velocity:brainly.com/question/13303224

#SPJ1

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Explanation:

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What is a an reason why a object gets extinct
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An object may become extinct due to the lack of quantity of the object and the object no longer exists so it is now extinct.

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A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge
Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance C_1 with energy of 7.54 J, is connected in parallel with another capacitor C_2 , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

U_O=q_0^2/2C_1=7.54 J\\

The energy stored in the electric field is the sum of the energies of the two capacitors:

U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2

since the charge equally distributed,  q_1 = q_2 = q_o/2. and since they are connected in parallel the potential difference on both of them is the same :

V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3\\

hence,

U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J

8 0
3 years ago
Desde el balcón de un edificio se deja caer una manzana y llega a la planta baja en 5 s. ¿Desde qué piso se dejó caer, si cada p
miv72 [106K]

Answer: 42

Explanation:

I will answer this in English.

We know that the apple needs 5 seconds to reach the ground.

Each floor of the building has a height of 2.88m.

Now, when we drop something, the only force acting on the object is the gravitational one, so the acceleration of the apple is:

a = -g

for the velocity, we integrate the acceleration over time, and as the apple is dropped, we do not have any initial velocity, so we do not have a constant in the integration:

v = -g*t

for the position we integrate again, now we have an initial height H, so the position is:

p = (-g/2)*t^2 + H

now the apple hits the ground when p = 0, so we can solve this equation to find H.

i will use g = 9.8m/s^2

0 = (-4.9m/s^2)*(5s)^2 + H  

H = 122.5 m

now knowing H, we can divide it by the height of a floor in the building and get the number of the floor.

N = 122.5m/2.88m = 42.5

this means that the apple was dropped in the floor 42 (the 0.5 means that the apple was not right where the floor 42 starts, it was dropped around the middle of the floor 42)

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3 years ago
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