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Afina-wow [57]
3 years ago
7

a diver on a board 7.5m above the water walks off the end with a horizontal velocity of 2.3 m/s when they hit the surface of the

water how far away from the edge of the board will they land ?
Physics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

2.846m

Explanation:

The diver is performing projectile motion.

To find x(final), we are going to use the equation x(final) = v(initial)*t + x(initial)

x(initial) = 0

x(final) = ?

v(initial) = 2.3 m/s

we don't know t

To find t we will use y(final) = 1/2*(-9.8)*t^2 + v(initial in the y dir.)*t + y(initial)

- 9.8 in the acceleration in the y dir.

y(final) = 0

y(initial) = 7.5

v(initial in the y dir.) = 0

If we solve for t we get: t = 1.237s

Now we have all the components to solve for x(final) in x(final) = v(initial)*t + x(initial)

x(final) = 2.3*1.237 + 0

x(final) = 2.846m

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You have a horizontal grindstone (a disk) that is 86 kg, has a 0.33 m radius, is turning at 92 rpm (in the positive direction),
Rudiy27

Answer:

a) α = 0.338 rad / s²  b)   θ = 21.9 rev

Explanation:

a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque

    τ = I α

    fr r = I α

Now we write the translational Newton equation in the radial direction

    N- F = 0

    N = F

The friction force equation is

    fr = μ N

    fr = μ F

The moment of inertia of a saying is

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Let's replace in the torque equation

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    α = 0.338 rad / s²

b) let's use the relationship of rotational kinematics

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    0 = w₀² - 2 α θ

    θ = w₀² / 2 α

Let's reduce the angular velocity

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    θ = 9.634 2 / (2 0.338)

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Let's reduce radians to revolutions

    θ = 137.3 rad (1 rev / 2π rad)

    θ = 21.9 rev

7 0
3 years ago
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