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3 years ago

Como se juega batminton? Explica con tus propias palabras

Physics

1 answer:

ad-work [718]3 years ago

7 0

Answer:

el batminton se juega con raquetas y con un volante no con pelota a diferencia de el tenis, el bádminton tiene diferentes reglas como que no toque el área local y que pasar el bote al otro lado de la red para que toque suelo para poder marcar, se juega con dos personas por cada lado el encuentro se juega a 3 sets el que llegue a dos gana....

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If an object is moving eastward and slowing down, then the direction of its acceleration is

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C. Eastward. Acceleration is the change in speed so it can be a positive (speeding up) or negative (slowing down) acceleration

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2 years ago

A block of mass m = 2.0 kg slides head on into a spring of spring constant k = 260 N/m. When the block stops, it has compressed

nasty-shy [4]

Answer:

a) Ws = 2.548 J

b) Wf = 1.153 J

c) v = 1.923 m / s

Explanation:

a) The work done by the spring force

Ws = ½ * k * x²

Ws = ½ * 260 N/m *0.14² m

Ws = 2.548J

b) The increase in thermal energy can by find using

Et = Wf

Wf = µ * m *g * x

Wf = 0.42 * 2.0 kg *9.8 m/s² * 0.14m

Wf = 1.153 J

c) The speed just as the block reaches can by fin using

EK = Ws + Et

Ek = ( 2.548 + 1.153 ) J = 3.7 J

Ek = ½ * m * v²

v² = 2* Ek / m

v = √[2 * 3.7 J / 2.0 kg]

v = 1.923 m / s

8 0

3 years ago

a ball of 5 kg is moving towards the wall at 6 m/s. after a while, it hits the wall and rebounds back at 4 m/s in the opposite d

iris [78.8K]

Answer:-50 J

Explanation:

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3 years ago

A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What

kramer

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let $p_{1}$ = momentum of the 1st ball

$p_{2}$ = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

$(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}$

$(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}$

Since we are dealing with identical balls, all the m terms cancel out so we are left with

$(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} + (v_{2})_{f}\cos \theta_{2}$

Putting in the numbers, we get

$1.33 = (0.750) \cos(33.30) + (v_{2})_{f} \cos \theta_{2}$

$= > (v_{2})_{f} \cos \theta_{2} = 0.703$

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

$0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}$

$= > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30) = 0.412$

Taking the ratio of the sine equation to the cosine equation, we get

$\frac{ \sin \theta _{2}}{ \cos \theta_{2} } = \tan \theta_{2} = \frac{0.412}{0.703} = 0.586$

$\theta_{2} = { \tan}^{ - 1} (0.586) = 30.4$

Solving now for $(v_{2})_{f}$ ,

$(v_{2})_{f} = \frac{0.412}{ \sin(30.4) } = 0.815 \: \frac{m}{s}$

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3 years ago

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position

Illusion [34]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Place of particles

$X(t)=\int V_{x}(t)dt$

$=\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C$

$\to t=0\\\\ \to X(0)=2.3 \ m$

$\to X(0)=0+C\\\\ \to C=2.3\ m$

$\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\$

$= \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\ = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m$

In point b:

when $t=2.2 \ s$

the Particle velocity $(V)=2 \times 2.22 =9.68\ \frac{m}{s}$

In point c:

Calculating the Particle acceleration:

$\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2 =8.8 \ \frac{m}{s^2}$

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3 years ago