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nignag [31]
3 years ago
8

The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank

has a density of 1.0 g/mL. How far has the boat sunk into the water?
Physics
1 answer:
Paladinen [302]3 years ago
8 0

When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat

So here weight of the boat = Buoyancy force

let say boat is sunk by distance "h"

now we can say

F_b = \rho * V * g

F_b = 1000*0.10 * 0.08 * h * 9.8

now by above force balance equation we can write

m*g = F_b

0.040 * 9.8 = 1000 * 0.10 * 0.08 * h * 9.8

0.040 = 8h

h = 5 * 10^{-3} m

so boat will sunk by total 5 mm distance

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3 years ago
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A package is dropped from a helicopter moving upward at 15 m/s
daser333 [38]

The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

The velocity of the helicopter from which the package was dropped = 15 m/s

The time it takes the package to strike the ground = 12 seconds

The required parameter:

The height of the package from the ground when it was dropped

The kinematic equation of motion relating distance, s, time, t, acceleration due to gravity, g, initial velocity, u, and final velocity, v, is applied as follows;

The package continues the upward motion for some time, t₁, given as follows;

Upward motion of the package

v = u - g·t₁

v = 0 at highest point reached by the package;

Therefore;

0 = 15 m/s - 9.81 m/s²  × t₁

t₁ = 15 m/s/(9.81 m/s²) ≈ 1.5295022 seconds

The time the package takes to return to the initial starting point, t₂ = t₁

The time the package falls after returning to the point it was dropped, t₃, is given as follows;

t₃ = t - (t₂ + t₁) = t - 2 × t₁

∴ t₃ = 12 s - 2 × 1.5295022 s ≈ 8.940996 s

From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

s = 15× 8.940996  + (1/2) × 9.81 × 8.940996² = 526.22755346 ≈ 530

The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

Learn more about the kinematic equations of motion here:

brainly.com/question/16995301

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2 years ago
Which of the following consists of several billion stars and interstellar matter?
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Determine the value of the resultant and its location from O.<br>see attach image.​
xxTIMURxx [149]

Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L  

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L  

the magnitude of the resultant will be

F = w₀L  √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

     (√⅛)[d] = ½         [⅔L] + ¼      [⅔L/2] - ¼      [L - ⅓L/2]

     (√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

     (√⅛)[d] =      ⅓L  +    ⅟₁₂L     -  ¼L + ⅟₂₄L  

     (√⅛)[d] = 5L/24

               d = 5L/24 / (√⅛)

               d = 5√⅛L/3

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The energy of the wave decreases gradually
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