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3 years ago

9. From the densities below, decide whether each substance will sink or float in sea water. Sea water

Physics

2 answers:

tekilochka [14]3 years ago

8 0

Answer:

Gasoline will float

Asphalt will sink

Cork will float

Explanation:

Simply compare the value of each object's density to that of the sea water (1.025 g/ml). If the density of the object is less than that of the water, the object will float due to the buoyance force.

Contrarily, if the density of the object is larger than that of sea water, the object will sink.

Gasoline, with density 0.66 g/ml which is less than that of sea water, will float.

Gasoline, with density 1.2 g/ml which is more than that of sea water, will sink.

Cork, with density 0.26 g/ml which is less than that of sea water, will float.

myrzilka [38]3 years ago

7 0

Answer:

gas will float asphalt will sink a cork will foat

Explanation:

basicaly fsf or float sink float

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k0ka [10]

Answer:

20 ft

Explanation:

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3 years ago

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2 years ago

Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o

Verdich [7]

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance = d = 96 miles

Lets call Pascal velocity V in miles per hour

Now if he increases his velocity by 50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V = d/ t₁ ⇒ 1.5 * V = 96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 ) ⇒ V * (t₁ + 16 ) = 4*d ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁ (1)

V*t₁ + 16*V = 384 (2)

We solve from equation (1) t₁ = 64/V

And by substitution in equation (2)

V * (64/V) + 16* V = 384

64 + 16 *V = 384 ⇒ 16*V = 320 ⇒ V= 320/16

V = 20 miles /sec

6 0

3 years ago

Find the ratio of the diameter of iron to copper wire, if they have the same resistance per unit length (as they might in househ

Natasha_Volkova [10]

Answer:

The ratio of the diameter of iron to Cu is;

$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

Explanation:

R=(ρL)/A

R is resistance,
L is length,
A is area,
ρ is resistivity
d is diameter

from the question the two materials have the same resistance per unit length.

$\frac{R}{L}= \frac{p}{A}$

$\frac{R}{L}$ for iron = $\frac{R}{L}$ for copper

This means we can equate ρ/A for both materials.

$\frac{p_{Fe} }{A_{Fe} } =\frac{p_{Cu} }{A_{Cu} }$

re-arranging the equation we have,

$\frac{A_{Fe}}{A_{Cu} } =\frac{p_{Fe} }{ p_{Cu} }$

$A=\pi \frac{d^{2} }{4}$

$\frac{A_{Fe}}{A_{Cu} } =\frac{d^{2}{Fe} }{ d^{2}{Cu} }$

$\frac{d^{2}{Fe} }{ d^{2}{Cu} } =\frac{p_{Fe} }{ p_{Cu} }$

$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

5 0

3 years ago

Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b

givi [52]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The equivalent gravitational force is ~

$F \approx1.48\times 10 {}^{ - 7} \: \: N$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

We know that ~

$\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}$

where,

F = gravitational force

$m_1$ = mass of 1st object = 500 kg

$m_2$ = mass of 2nd object = 20kg

G = gravitational constant = $6.674 × {10}^ {-11}$

r = distance between the objects = 2.12 m

Let's calculate the force ~

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 500 \times 20}{(2.12) {}^{2} }$

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}$

$F = \dfrac{6.674}{4.4944} \times 10 {}^{ - 7}$

$F =1.484 \times 10 {}^{ - 7} \: \: newtons$

7 0

2 years ago