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3 years ago

A ball is thrown vertically up ward with a velocity of 25m/s. a/how fast was it moving after 2sec?

Physics

1 answer:

Shalnov [3]3 years ago

3 0

Answer:

12.5

Explanation:

given

distance=25

time=2sec

required

=speed

solution

=v=s/t

25/2

=12.5m/s

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2 years ago

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving alo

vazorg [7]

Answer with Explanation:

We are given that

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$m'=0.20 kg$

Initial velocity of second puck,u'=5.2m/s

Final velocity of second puck , $v'=3.1m/s$

$\theta=53^{\circ}$

a.According to law of conservation of momentum

Along x- axis

$mu_x+m'u'_x=mv_x+m'v'cos\theta$

Substitute the values

$0.30\times 0+0.20\times 5.2=0.30v_x+0.20\times 3.1cos53$

$1.04=0.3v_x+0.372$

$1.04-0.372=0.3v_x$

$0.668=0.3v_x$

$v_x=\frac{0.668}{0.3}=2.23m/s$

Along y- axis

$mu_y+m'u'_y=mv_y+m'v_ysin53$

$0.3\times 0+0.2\times 0=0.3\times v_y+0.2\times 3.1sin53$

$0=0.3v_y+0.495$

$-0.495=0.3v_y$

$v_y=\frac{-0.495}{0.3}=-1.65m/s$

$v=\sqrt{v^2_x+v^2_y}=\sqrt{(2.23)^2+(1.65)^2}=2.77m/s$

Hence, the velocity after the collision of the 0.3 lg puck after the collision=2.77 m/s

b.Initial kinetic energy= $E_i=\frac{1}{2}mu^2+\frac{1}{2}m'u'^2=0+\frac{1}{2}(0.2)(5.2)^2=2.704 J$

Final kinetic energy= $E_f=\frac{1}{2}mv^2+\frac{1}{2}m'v'^2=\frac{1}{2}(0.3)(2.77)^2+\frac{1}{2}(0.2)(3.1)^2=2.1 J$

Fraction of kinetic energy lost in the collision= $(1-\frac{E_f}{E_i})\times 100$

Fraction of kinetic energy lost in the collision= $(1-\frac{2.1}{2.704})\times 100$ =22.34%

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3 years ago

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A ball is thrown vertically up ward with a velocity of 25m/s. a/how fast was it moving after 2sec?​

A ball is thrown vertically up ward with a velocity of 25m/s. a/how fast was it moving after 2sec?