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sladkih [1.3K]
3 years ago
7

A stationary, 32 kilogram ice skater catches a 0.5 kilogram snowball thrown at a speed of 45 meters per second. The collision is

totally inelastic. How fast, in meters per second, does the skater (and snowball) move afterward? Friction of the skater on the ice is negligible.
Physics
1 answer:
Pie3 years ago
4 0

Answer:

V = 0.69 m/s

Explanation:

Given that,

The mass of a ice skater, m₁ = 32 kg

The initial speed of the ice skater, u₁ = 0

Mass of a snowball, m₂ = 0.5 kg

The initial speed of a snowball, u₂ = 45 m/s

eIt is mentioned that the collision is totally inelastic. Let V be the common speed of skater and the ball. So, using the law of conservation of momentum.

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\\V=\dfrac{32\times 0+0.5\times 45}{32+0.5}\\\\V=0.69\ m/s

So, the common speed is equal to 0.69 m/s.

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The optical fiber has two layers, one is called core and the other is called cladding.

The refractive index of core is more than the cladding.

As a ray of light falls on the core at more than the critical angle for that pair of media(core and cladding), it suffers the total internal reflection at the interface.

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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th
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Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2

The angular acceleration when it stopping:

\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2

The angular distance it covers when starting from rest:

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\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad

The angular distance it covers when coming to complete stop:

0 - \omega^2 = 2\alpha_o\theta_o

\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

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If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step
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Add both sides by x_i:

x_i+\frac{v_i^2 - v_f^2}{2a} =x_f

Subtract both sides by \frac{v_i^2 - v_f^2}{2a}:

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