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3 years ago

I have a question, it concerns hydrostatic buoyancy and Archimedes' law.

Physics

1 answer:

saw5 [17]3 years ago

4 0

Answer: the lvl wud remain the same

Explanation: as per Archimedes Principle, the weight of the water displaced by the object is equal to the weight of the object. When the ship initially went into the pool, it wud hv displaced some water. When the anchor is dropped, the level does not change coz the anchor was already in the ship and no extra weight has been added, so the weight of the anchor has already been accounted for in the first place when the ship was first placed in the pool

You might be interested in

What is the current I(3τ), that is, the current after three time constants have passed? The current in the circuit will approach

Olin [163]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$I(\tau)=0.051 A$

$I(3 \tau)=0.076 A$

$I_c= 0.08 A$

Explanation:

From the question we are told that

$I(t) = \frac{e}{R}(1-e^{\frac{t}{\tau} }) ; \ Where \ \tau = L/R$

From the question we are told to find $I(\tau)$ when t=0 equals the time constant ( $\tau$ )

That is to obtain $I(\tau)$ .This is mathematically represented as

$I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{\tau}{\tau} })$

Substituting 12 V for $\epsilon$ and 150Ω for R

$I(\tau) = \frac{12}{150} (1- e^{-1})$

$=0.051 A$

From the question we are told to find $I(3 \tau)$ when t=0 equals the 3 times the time constant ( $\tau$ )

That is to obtain $I(3\tau)$ .This is mathematically represented as

$I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{3\tau}{\tau} })$

$I(\tau) = \frac{12}{150} (1- e^{-3})$

$=0.076 A$

As tends to infinity $\frac{\infty}{\tau} = \infty$

So $I_c$ would be mathematically evaluated as

$I_c=I(\infty) = \frac{12}{150} (1- e^{- \infty})$

$= \frac{12}{150}$

$= 0.08 A$

5 0

3 years ago

The mercalli scale is a scale from ________.

Mashcka [7]

<h2>Answer: 1 to 12. </h2>

The Mercalli seismological scale is a scale of 12 degrees, where 1 is very weak and 12 catastrophic.

This scale owes its name to the Italian physicist Giuseppe Mercalli and was developed to assess the intensity of earthquakes through the effects and damages caused to different structures.

This means the Mercalli scale measures how strong a earthquake has been by its consequences and not by its magnitude, therefore it is based on empirical observations.

6 0

3 years ago

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

nevsk [136]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

4 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reacti

Yakvenalex [24]

The equilibrium constant of the reaction at 25⁰c will be 426827.5.

Theory-

Equilibrium constant :The equilibrium constant comes from the chemical equilibrium law. For the chemical equilibrium state, at a fixed constant temperature, the ratio of the product of the reaction's multiplication to the concentration of its reactants' multiplication, and each is raised to the power to the corresponding coefficients of the elements in the reaction.

The chemical equilibrium is given by for a general chemical reaction.

a. A+ b. B ⇌ c. C+ d. D,.

Kc =[C]c [D]d/[A]a [B]b.

Gibb's free energy :The second law of thermodynamics can be arranged in such a way that it gives a new expression when a chemical reaction happens at a constant temperature and constant pressure.

G=H-TS

Calculations:-

T=25⁰c

G=51.4 x 10³J

$\\\\k=GR+\frac{nRT}{Z} \\$

k= equilibrium constant ,G=Gibbs free energy ,n= no. of moles ,R=Gas constant ,T=temperature ,Z=compressibility

$Ideal.Situation=\left \{ {{Z=1} \atop {n=1}} \right.$

$\\\\\\k=GR+RT$

k=51.4 x 10³ x 8.3 + 8.3 x 25

k=426827.5

To learn equilibrium constant-

brainly.com/question/19669218

#SPJ4

6 0

1 year ago

A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f

Setler [38]

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

$W_f = F_fs = 15*0.5 = 7.5 J$