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yuradex [85]
3 years ago
12

Enter an expression for the force constant for the floating raft, in terms of L, g, and the density of water, ρ.

Physics
1 answer:
andriy [413]3 years ago
3 0

Answer:

K = ρL²g

Explanation:

Consider L as the length of the raft inside the water when the raft is displaced through additional distance y;

Then:

F = upthrust ( restoring force) = weight of the liquid displaced.

F = V_{\omega} \rho_{\omega} g= A y \rho_{\omega} g

where;

A = L²

\rho_{\omega} = \rho

F = ky.

Then,

Ay \rho g = ky

L^2y \rho g = ky

Divide both sides by y

K = ρL²g

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You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa
rusak2 [61]

The distance you free-fall from rest is  D = (1/2) (g) (T²) <== memorize this

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Height = (4.9/5.76) meters

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3 years ago
A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop
postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination(\theta )=30^{\circ}

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

F_{net}=0

Force applied by cable=mg\sin \theta

F=60\times 9.8\times \sin (30)=294 N

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The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
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Answer:

velocity = 1527.52 ft/s

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Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

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Answer:

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