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2 years ago

14

A 97 kg man lying on a surface of negligible friction shoves a 62 g stone away from himself, giving it a speed of 2.6 m/s. What

speed does the man acquire as a result?

1 answer:

tangare [24]2 years ago

3 0

Answer:

man will move in opposite direction with speed

$v_1 = 1.66 \times 10^{-3} m/s$

Explanation:

As we know that man is lying on the friction-less surface

so here net force along the surface is zero

so if we take man + stone as a system then net change in momentum of this system will become zero

so here we have

$P_i = P_f$

$0 = m_1v_1 + m_2v_2$

here we have

$0 = (97)v_1 + 0.062(2.6)$

$v_1 = -\frac{0.1612}{97}$

$v_1 = -1.66 \times 10^{-3} m/s$

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pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations:

$\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N$ (2)

$\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N$ (3)

The solution of this system is:

$F_{3,x} = 56.148\,N$ , $F_{3,y} = -116.884\,N$

The x-component of $F_{3}$ is 56.148 newtons.

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2 years ago

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

worty [1.4K]

Answer:

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Explanation:

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$\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h$

The minimum velocity is:

$v_{min} = \sqrt{2\cdot g \cdot h}$

Let assume that radio of curvature is measured in meters. Hence:

$v_{min} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(15\,m)}$

$v_{min} \approx 17.153\,\frac{m}{s}$

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2 years ago

a stone is dropped from the top of 50 m high tower simultaneously another stone is thrown upward with a speed of 20 m/s . calcul

Darina [25.2K]

'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.

For the stone dropped from the 50m tower:

H = +50 - (1/2) G T²

For the stone tossed upward from the ground:

H = +20T - (1/2) G T²

When the stones' paths cross, their <em>H</em>eights are equal.

50 - (1/2) G T² = 20T - (1/2) G T²

Wow ! Look at that ! Add (1/2) G T² to each side of that equation,
and all we have left is:

50 = 20T Isn't that incredible ? ! ?

Divide each side by 20 :

<u>2.5 = T</u>

The stones meet in the air 2.5 seconds after the drop/toss.

I want to see something:
What is their height, and what is the tossed stone doing, when they meet ?

Their height is +50 - (1/2) G T² = 19.375 meters

The speed of the tossed stone is +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it. The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.

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