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dangina [55]
2 years ago
5

En las olimpiadas del 2012 del colegio villapalos maria gano la carrera de los 100 m en 10,56 s y la de 200 m en 22,34 s ¿en cua

l fue mas veloz
Physics
1 answer:
daser333 [38]2 years ago
6 0

Answer:

Fue mas velóz en los 100 m.

v_{1}=9.47\: m/s

Explanation:

Tomando en cuanta que la definición de velocidad es:

v=\frac{\Delta x}{t}

Donde:

  • Δx es el desplazamiento
  • t es el tiempo

La primera velodcidad esta dada por:

v_{1}=\frac{100}{10.56}

v_{1}=9.47\: m/s

La segunda velocidad sera:

v_{1}=\frac{200}{22.34}

v_{1}=8.95\: m/s

Por lo tanto, en el primer tramo fue más rápido.

Espero te haya ayudado!

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Elena L [17]

Answer:

yes

Explanation:

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Which of the following methods of improving air quality does not require advanced technology?
Kisachek [45]
Conservation is a method of improving air quality
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a load of 400 Newton is lifted by a first class lever in which the load is at the distance of 20 cm and the effort is at the dis
Len [333]

Answer:

  1. solution,
  2. Given
  3. load =400N
  4. ld=0.2m
  5. ed=0.6m
  6. effort =150N

Explanation:

efficiency =output work/input work ×100%

l×ld/e×ed×100%

400×0.2/150×0.6×100%

80/90×100%

88.89%ans

7 0
3 years ago
A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the
Shalnov [3]

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

8 0
3 years ago
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0
3 years ago
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