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2 years ago

A hooligan throws a stone vertically down with an initial speed of 17 m/s from the roof of a building 51 metres above the

Physics

1 answer:

pshichka [43]2 years ago

4 0

Answer:

Explanation:

impact velocity is

√(18² + 2(-9.8)-51) = 36 m/s

average velocity is (17 + 36)/2 = 26.5 m/s

t = d/v = 51 / 26.5 = 1.9 s

if ground is origin and UP positive

0 = 51 - 17t - ½(9.8)t²

quadratic formula

t = (17 ±√(17² - 4(-4.9)(51))) / (2(-4.9))

t = (17 ± 36) / -9.8

t = - 5.4 s which we ignore as the stone is not thrown yet

t = 19 s

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Why are different colors used to represent the floor of the ocean?

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The red, yellow, and green wavelengths of sunlight are absorbed by water molecules in the ocean. ... In coastal areas, runoff from rivers, resuspension of sand and silt from the bottom by tides, waves and storms and a number of other substances can change the color of the near-shore waters.

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2 years ago

After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg

GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses $m_1$ and $m_2$ whose respective velocities before collision are $u_1$ and $u_2$ ;

$m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)$

where $v_1$ and $v_2$ are their respective velocities after collision.

Given;

$m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s$

Note that $u_2$ =0 because the second mass $m_2$ was at rest before the collision.

Also, since the two masses are equal, we can say that $m_1=m_2=m$ so that equation (1) is reduced as follows;

$mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)$

m cancels out of both sides of equation (2), and we obtain the following;

$u_1+u_2=v_1+v_2.............(3)$

a) When $v_1=0.8m/s$ , we obtain the following by equation(3)

$5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s$

b) As $m_1$ stops moving $v_1=0$ , therefore,

$5+0=0+v_2\\v_2=5m/s$

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3 years ago

A 5.00 kg pendulum swings back and forth. At the top of its arc it reaches a height of 0.36 m. What is the velocity of the pendu

azamat

Answer:

Gravitational potential energy = (mass) x (gravity) x (height)

Kinetic energy (of a moving object) = (1/2) (mass) x (speed)²

When the pendulum is at the top of its swing, its potential energy is

(mass) x (gravity) x (height)

= (5 kg) x (9.8 m/s²) x (0.36 m)

= (5 x 9.8 x 0.36) joules

17.64 joules .

Energy is conserved ... it doesn't appear or disappear ... so that number is exactly the kinetic energy the pendulum has at the bottom of the swing, only now, it's kinetic energy:

17.64 joules = (1/2) x (mass) x (speed)²

17.64 joules = (1/2) x (5 kg) x (speed)²

Divide each side by 2.5 kg:

17.64 joules / 2.5 kg = speed²

Write out the units of joules:

17.64 kg-m²/s² / 2.5 kg = speed²

(17.64 / 2.5) (m²/s²) = speed²

7.056 m²/s² = speed²

Take the square root of each side: Speed = √(7.056 m²/s²) = 2.656 m/s.

Looking through the choices, we're overjoyed to see that one if them is ' 2.7 m/s '. Surely that's IT !

Hope this will help you ...

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