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natulia [17]
3 years ago
8

Sonia O'Sullivan of Ireland set the World Record in 1994 for the Women's 1000 m race with a time of 2 minutes and 45 seconds. Wh

at was her average velocity?​ ill give brainliest
Physics
2 answers:
Aleks04 [339]3 years ago
5 0

Answer:

Sonia O'Sullivan (born 28 November 1969) is an Irish former track and field athlete. She won a gold medal in the 5000 metres at the 1995 World Championships, and a silver medal in the 5000 metres at the 2000 Olympic Games.[1] Her 2000 m world record of 5:25.36, set in 1994 stood until 2017.[2]

O'Sullivan first came to prominence when winning the 1500 m at the 1991 Universiade, before going on to finish fourth in the 3000 m final at the 1992 Olympic Games. She then won a silver medal in the 1500 m at the 1993 World Championships. She was the favourite for the 5000 m title at the 1996 Olympic Games but dropped out of the final due to illness. As well as her 1995 World title, she won three gold medals at the European Championships, in the 3000 m (1994), 5000 m (1998) and 10,000 m (1998), and is a two-time World Cross Country Champion.

O'Sullivan won silver medals in the 5000 m and 10,000 m at the 2002 European Championships, and competed at her fourth Olympic Games in 2004. She is known for her dramatic kick, clocking 28-second final 200 m splits in some of her races.[3]

She is the one of two women (the other Tirunesh Dibaba) who won the short and long course World Cross Country title at the same championship (1998 in Marrakesh).[4]

you can find from here

puteri [66]3 years ago
5 0

Answer:

6.06061 m/s

Explanation:

1000 m in 165sec

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An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t
marissa [1.9K]
The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
where 
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
4 0
3 years ago
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block
kotegsom [21]

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  v  = 11.76  \ m/ s

Explanation:

From the question we are told that

   The  total distance traveled is  d =  1.2  \ m

    The mass of the block is  m_b  =  0.3 \ kg

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   PE  =  KE

Where  PE  is the potential energy which is mathematically represented as

      PE  =  m * g  *  h

substituting values

     PE  =   3 *  9.8  *  0.60

      PE  =  17.64 \  J

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          KE  =  \frac{1}{2}  *  m v^2

So

      \frac{1}{2}  *  m* v ^2  =  PE

substituting values  

  =>    \frac{1}{2}  *  3 * v ^2  = 17.64

=>       v  = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }

=>    v  = 11.76  \ m/ s

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Goryan [66]

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4 0
3 years ago
What is the magnitude of the force required to stretch a 20 cm-long spring, with a spring constant of 100 N/m, to a length of 21
GalinKa [24]

The correct answer to the question is: 1 N.

EXPLANATION:

As per the question, the spring constant or the force constant of the spring is given as k = 100 N/m.

The original length of the spring L = 20 cm.

The stretched length of the spring L'= 21 cm.

Hence, the change in length will be-

                              ∆L = L' - L

                                    = 21 cm - 20 cm

                                    = 1 cm

                                    = 0.01 m

We are asked to calculate the magnitude of force acting on the spring .

From Hooke's  law, we know that the restoring force that acts on the spring is proportional to the distance .

Mathematically it can be written as -

                F = - kx.

Here, k is the force constant.

         x is the change in length due to compression or elongation.

The negative sign is due to the fact that it is opposite to the applied force.


Hence, the applied force on the spring is calculated as -

            F = kx

               = k × ∆L

               = 100 N/ m × 0.01 m

               = 1 N.

Hence, the force acting on the spring is 1 N.


                                                   

8 0
3 years ago
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