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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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ladessa [460]

1).  <u>Power = (voltage)² / (Resistance)</u>

     4,500 = (220)² / Resistance

Multiply each side by (resistance) :  4,500 x resistance = (220)²

Divide each side by  4,500 :            Resistance  =  (220)² / 4,500 = <em>10.76 ohms</em>


2).  <u>Power = (voltage) x (Current)</u>

Divide each side by (voltage):  Power / voltage = Current

                                            4,500 / 220  =  <em>20.45 Amperes</em>


3).  4,500 watts = 4.5 kilowatts

     (4.5 kilowatts) x (4 hours)  =  <em>18 kilowatt-hours</em>


3 0
2 years ago
Fill in the blanks to complete the statements.
Murljashka [212]

Answer:

When an object changes speed (increases/decreases) it results in acceleration/de acceleration, its velocity also changes.

Explanation:

Acceleration is the rate of change in velocity.An object can accelerate when speed increases, decreases or direction changes. All these instances involves a change in velocity.Velocity is a vector quantity thus it has magnitude and the direction.Acceleration due to change in direction is centripetal acceleration.The expression for finding acceleration is;

a=change in velocity/change in time

a=Δv/Δt in m/s²

3 0
2 years ago
A 0.900 kg ornament is hanging by a 1.50 m wire when the ornament is suddenly hit by a 0.400 kg missile traveling horizontally a
just olya [345]

Explanation:

The given data is as follows.

  Mass of the ornament (m_{1}) = 0.9 kg

  Length of the wire (l) = 1.5 m

 Mass of missile (m_{2}) = 0.4 kg

 Initial speed of missile (u_{2}) = 12 m/s

         r = 1.5 m

According to the law of conservation of momentum,

                   p_{i} = p_{f}  

     m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v

Putting the given values into the above formula as follows.

          m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v

         0.9 \times 0 + 0.4 \times 12 = (0.9 + 0.4)v

              0 + 4.8 = 1.3v

                  v = 3.69 m/s

Now, the centrifugal force produced is calculated as follows.

            F_{c} = (m_{1} + m_{2}) \times \frac{v^{2}}{r}

                       = (0.9 + 0.4) \times \frac{(3.69)^{2}}{1.5}

                       = 11.80 N

Hence, tension in the wire is calculated as follows.

              T = F_{c} + (m_{1} + m_{2})g

                 = 11.80 N + (0.9 + 0.4) \times 9.8

                 = 24.54 N

Thus, we can conclude that tension in the wire immediately after the collision is 24.54 N.

4 0
2 years ago
A 3.0-cm tall statue is 48 cm in front of a convex lens having a focal length of i. (2 pts) Is the image of the statue real or v
erica [24]

Explanation:

Given that,

Height of object = 3.0 cm

Distance of object u= 48 cm

Focal length = 48 cm

We need to calculate the image distance

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{-48}

\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{48}

\dfrac{1}{v}=\dfrac{17}{240}

v=14.11\ cm

(I). The image is real.

(II).  The distance of the image from the lens is 14.11 cm.

(II). The image is inverted.

(IV). We need to calculate the height of the image

Using formula of magnification

m = \dfrac{v}{u}=\dfrac{h'}{h}

\dfrac{v}{u}=\dfrac{h'}{h}

Put the value into the formula

\dfrac{14.11}{48}=dfrac{h'}{3.0}

h'=\dfrac{14.11}{48}\times3.0

h'=0.88\ cm

The height of the image is 0.88 cm.

Hence, This is the required solution.

6 0
2 years ago
A solenoid 4.0 cm in radius and 4.0 m in length has 8000 uniformly spaced turns and carries a current of 5.0 A. Consider a plane
UkoKoshka [18]

Answer:

Magnetic flux, \phi=1.57\times 10^{-5}\ Wb

Explanation:

It is given that,

Radius of the solenoid, r = 4 cm = 0.04 m

Length of the solenoid, L = 4 m

No of turns, N = 8000

Current, I = 5 A

Radius of the plane circular surface, r' = 2 cm = 0.02

Area of the circular surface,

A=\pi r'^2

A=\pi (0.02)^2=0.00125\ m^2                  

The magnetic flux through this surface is given by :

\phi=B\times A

B is the magnetic field of the solenoid

\phi=\mu_o\dfrac{N}{L}I\times A

\phi=4\pi \times 10^{-7}\times \dfrac{8000}{4}\times 5\times 0.00125

\phi=1.57\times 10^{-5}\ Wb

So, the magnetic flux through this surface is 1.57\times 10^{-5}\ Wb/. Hence, this is the required solution.

3 0
2 years ago
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