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3 years ago

What is the horizontal component of a ball thrown at a 27 degree angle at 16 m/s?

Physics

1 answer:

Airida [17]3 years ago

6 0

Answer:

14.25 m/s

Explanation:

In this problem, we need to find the horizontal component of a ball thrown at a 27 degree angle at 16 m/s.

It can be given by :

$v_x=v\cos\theta\\\\=16\times \cos(27)\\\\=14.25\ m/s$

So, the horizontal component of the ball is 14.25 m/s.

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3 years ago

A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o

vredina [299]

Answer:

Recoil speed, $1.17\times 10^{-5}\ m/s$

Explanation:

Given that,

Mass of the comet fragment, $m_1=1.96\times 10^{13}\ kg$

Speed of the comet fragment, $v_1=6.5\times 10^4\ m/s$

Mass of Callisto, $m_2=1.08\times 10^{23}\ kg$

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$m_1v_1=(m_2+m_2)V$

V is recoil speed of Callisto immediately after the collision.

$V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s$

So, the recoil speed of Callisto immediately after the collision is $1.17\times 10^{-5}\ m/s$

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3 years ago

Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer

sveticcg [70]

by cosine law we know that

$c^2 = a^2 + b^2 - 2 abcos\theta$

$\theta = 180 - 54.3 - 79.1 = 46.6 degree$

now using above equation

$93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6$

$4942.03 = b^2 - 85.34 b$

$b^2 - 85.34b - 4942.03 = 0$

by solving above quadratic equation we have

$b = 124.9 m$

so it is at distance 124.9 m from deer a

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3 years ago