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Setler79 [48]
2 years ago
5

A gymnast practices two dismounts from the high bar on the uneven parallel bars. during one dismount, she swings up off the bar

with an initial upward velocity of + 4.0 m/s. in the second, she releases from the same height but with an initial downward velocity of −3.0 m/s. what is her acceleration in each case? how do the final velocities of the gymnast as she reaches the ground differ?
Physics
1 answer:
Fiesta28 [93]2 years ago
8 0
Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.

g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.

First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s

Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s

The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.

Answer:
First dismount:
  Acceleration  = 9.8 m/s² downward
  Landing velocity = 8.42 m/s downward

Second dismount:
  Acceleration = 9.8 m/s² downward
  Landing velocity = 7.99 m/s downward

The landing velocities differ by 0.43 m/s.

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(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0
2 years ago
An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and
torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

 

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

 

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552  ; from this we get AC = 172 km (3 significant figures)

 

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

 

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

 

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

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2 years ago
The planets how and block are near each other in the Dorgon system. the Dorgons have very advanced technology, and a Dorgon scie
BlackZzzverrR [31]

Answer:

Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox.

Explanation:

The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.

Or

If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.  

On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.

4 0
2 years ago
It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 110 m above a raging riv
zzz [600]

Answer:

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Explanation:

First we need the speed when the cord starts stretching:

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V_2^2=-2*10*(-31)

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By conservation of energy:

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h is your height at its maximum elongation

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21h^2-2518h+43060.6=0 Solving for h:

h_1=20.66m  and h_2=99.24m  Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

6 0
2 years ago
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vivado [14]
Specific\ resistance\ =resistivity\\
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3 0
2 years ago
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