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Alex_Xolod [135]
4 years ago
14

Three ice skaters meet at the center of a rink and each stands at rest facing the center, within arm's reach of the other two. O

n a signal, each skater pushes himself away from the other two across the frictionless ice. After the push, skater A with mass mA = 70.0 kg moves in the negative y-direction at 2.00 m/s and skater B with mass m = 85.0 kg moves in the negative x-direction at 2.50 m/s. Find the x- and y-components of the 80.0 kg skater C's velocity (in m/s) after the push. HINTvcx= ___ m/svcy=____m/s
Physics
1 answer:
icang [17]4 years ago
5 0

To solve the problem it is necessary to apply the conservation equations for the moment, specifically for collision. In addition to that, the concepts of vector velocity are necessary, in which the components are obtained and if the total magnitude is necessary.

Conservation of Momentum equation is given by,

m_1v_1+m_2v_2+m_3v_3 = 0

We need to find the speed 3, therefore by readjusting the equation we have,

v_c = \frac{-(m_1v_1+m_2v_2)}{m_3}

v_c = \frac{-(-70*2\hat{j}-85*2.5\hat{j})}{80}

v_c = 2.656\hat{i}+ 1.75\hat{j}

Therefore the components of the velocity would be,

v_{cx} = 2.656m/s

v_{cy} = 1.75m/s

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A climber using bottled oxygen accidentally drops the oxygen bottle from an altitude of 4500 m. If the bottle fell straight down
Trava [24]

Answer:

v= 300 m/s

Explanation:

Given that

altitude ,h= 4500 m

The mass ,m = 3 kg

Lets take acceleration due to gravity , g= 10 m/s²

The speed before impact at sea  level =  v

Initial speed ,u = 0 m/s

We know that

v²=u²+2 g h

v=final speed

u=initial speed

h=height

Now by putting the values in the above equation

v² = 0²+ 2 x 10 x 4500

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v= 300 m/s

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3 0
3 years ago
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
blondinia [14]

Answer:

His pitching speed is 38 m/s.

Explanation:

Hi there!

Please see the attached figure for a better understanding of the problem.

The position of the ball at any time t is given by the following vector:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector of the ball at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference at the throwing point so that x0  and y0 = 0.

When the ball reaches the ground, its position vector will be r1 (see figure). Using the equation of the vertical component of the position vector, we can find the time at which the ball reaches the ground. At that time, the horizontal component of the position is 30 m and the vertical component is -3.0 m (see figure):

y = y0 + 1/2 · g · t²  (y0 = 0)

y = 1/2 · g · t²

-3.0 m = 1/2 · (-9.8 m/s²) · t²

-3.0 m / -4.9 m/s² = t²

t = 0.78 s

Now, knowing that at this time x = 30 m, we can find v0:

x = x0 + v0 · t  (x0 = 0)

x = v0 · t

30 m = v0 · 0.78 s

v0 = 30 m / 0.78 s

v0 = 38 m/s

His pitching speed is 38 m/s.

3 0
4 years ago
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