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2 years ago

A white pool ball of mass 1.0 kg moving at 10 m/s collides with

Physics

1 answer:

solniwko [45]2 years ago

7 0

The velocity of the red ball after the collision is 5.8 m/s

Explanation:

In absence of external forces on the system, we can apply the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 1.0 kg$ is the mass of the pool ball

$u_1 = 10 m/s$ is the initial velocity of the pool ball

$v_1 = 3.0 m/s$ is the final velocity of the pool ball

$m_2 = 1.2 kg$ is the mass of the red ball

$u_2 = 0$ is the initial velocity of the red ball

$v_2$ is the final velocity of the red ball

Solving the equation for v2, we find the final velocity of the red ball after the collision:

$v_2 = \frac{m_1 u_1-m_1v_1}{m_2}=\frac{(1.0)(10)-(1.0)(3.0)}{1.2}=5.8 m/s$

Learn more about collisions:

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010) Identify the true statement. Group of answer choices The height of waves is determined by wind strength and fetch. Wave bas

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Answer:

The height of the wave is determined by the wind strength and fetch.

Explanation:

The height of the wave is determined by the wind strength and fetch.

The more the strength and the more the fetch size the more will be the height of the wave.

Remember as the wave approaches the coast its wavelength decreases and the wave height increases, whereas when the wave goes away from the coast its wavelength increases and height decreases.

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2 years ago

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th

zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(d) K = 11.8835 J

U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)

⇒ f = 1.3368 Hz

we use the formula

E = m*ω²*A² / 2

⇒ E = (1.15)(8.40)²(0.650)² / 2

⇒ E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)

and

U = (1/2)*m*ω²*x² (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒ K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒ U = 5.2581 J

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2 years ago

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

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To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed