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2 years ago

Buoyancy increases with the increase in the density of a) Submerged body b) Fluid

Physics

1 answer:

AlekseyPX2 years ago

4 0

Answer:

Submerged body

Explanation:

If buoyancy is greater than weight then object will float.
If buoyancy is less then weight then object will sink.
If buoyancy=weight then objects remains stable

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A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect

zaharov [31]

Answer:

$X=X_o+\dfrac{1}{2}gt^2$

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

$\dfrac{1}{2}mV^2=mg(X-X_o)$

$V=\sqrt{2g(X-X_o)}$

We know that

$\dfrac{dX}{dt}=V$

$\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}$

$\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX$

At t= 0 ,X=Xo

So we can say that

$X=X_o+\dfrac{1}{2}gt^2$

So the length of cable after time t

$X=X_o+\dfrac{1}{2}gt^2$

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2 years ago

What is one use for infrared waves? A radar B Medical imaging C Thermal imaging cameras

Vilka [71]

Answer:

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Explanation:

...

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2 years ago

Which of the following is considered a physical change? burning paper rusting of metal evaporating water

alex41 [277]

When paper burns, some of the chemical compounds in the
paper combine with oxygen in the air and become different
chemical compounds. That's a chemical change.

When iron rusts, or copper or silver turns green, that's the result
of the metal at the surface combining with the oxygen in the air and
forming a new chemical compound. Those are chemical changes.

When water evaporates, H₂O in the liquid phase gains thermal
energy and changes to H₂O in the gaseous phase. No chemical
compounds are lost, gained, or changed to other compounds.
It's just a physical change.

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2 years ago

An object experiences a constant change in velocity in free fall.<br><br> True False

jeyben [28]

False, it experiences a constant change in ACCELERATION in free fall.

3 0

2 years ago

A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 40

11111nata11111 [884]

Quasi frequency = 4√6

Quasi period = π√6/12

t ≈ 0.4045

<u>Explanation:</u>

Given:

Mass, m = 20g

τ = 400 dyn.s/cm

k = 3920

u(0) = 2

u'(0) = 0

General differential equation:

mu" + τu' + ku = 0

Replacing the variables with the known value:

20u" + 400u' + 3920u = 0

Divide each side by 20

u" + 20u' + 196u = 0

Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.

r² + 20r + 196 = 0

Determining the roots:

$r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}$

r = -10 ± 4√6i

The general solution for two complex roots are:

y = c₁ eᵃt cosbt + c₂ eᵃt sinbt

with a the real part of the roots and b be the imaginary part of the roots.

Since, a = -10 and b = 4√6

u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t

u(0) = 2

u'(0) = 0

(b)

Quasi frequency:

μ = $\frac{\sqrt{4km - y^2} }{2m}$

$= \frac{\sqrt{4(3929)(20) - (400)^2} }{2(20)} \\\\= 4\sqrt{6}$

(c)

Quasi period:

T = 2π / μ

$T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6} }{12}$

(d)

|u(t)| < 0.05 cm

u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05

solving for t:

τ = t ≈ 0.4045

8 0

2 years ago

Buoyancy increases with the increase in the density of a) Submerged body b) Fluid​

Buoyancy increases with the increase in the density of a) Submerged body b) Fluid