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3 years ago

If a substance has a density of 2.7g/cm3 and a mass of 86.4g, what is its volume?

Physics

1 answer:

Gennadij [26K]3 years ago

7 0

Answer:

32cm³

Explanation:

Given parameters:

Density of substance = 2.7g/cm³

Mass of substance = 86.4g

Unknown:

Volume of substance = ?

Solution:

Density is the mass per unit volume of a substance.

Density = $\frac{mass}{volume}$

Since the unknown is volume we solve for it;

mass = density x volume

86.4 = 2.7 x volume

volume = $\frac{86.4}{2.7}$ = 32cm³

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Nina [5.8K]

Answer:

Acceleration = 0.0282 m/s^2

Distance = 13.98 * 10^12 m

Explanation:

we will apply the energy theorem

work done = ΔK.E ( change in Kinetic energy ) ---- ( 1 )

where :

work done = p * t

= 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J

( note : convert 1 year to seconds )

and ΔK.E = 1/2 mVf^2 given ; m = 1200 kg and initial V = 0

back to equation 1

473040000 * 10^6 = 1/2 mv^2

Vf^2 = 2(473040000 * 10^6 ) / 1200

∴ Vf = 887918.92 m/s

i) Determine how fast the rocket is ( acceleration of the rocket )

a = Vf / t

= 887918.92 / ( 1 year )

= 0.0282 m/s^2

ii) determine distance travelled by rocket

Vf^2 - Vi^2 = 2as

Vi = 0

hence ; Vf^2 = 2as

s ( distance ) = Vf^2 / ( 2a )

= ( 887918.92 )^2 / ( 2 * 0.0282 )

= 13.98 * 10^12 m

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3 years ago

How do streams change as they flow from mountains down to plains? Plz help URGENT I need it today

lisov135 [29]

They start as one stream, then get curvier, then empy into a delta. A delta is a ton of streams of water. It is one, then empties out into a ton of streams.
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3 years ago

A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the

ivolga24 [154]

Answer:

The acceleration of the wallet is $3\hat{i}+6\hat{j}$

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse $a=2\hat{i}+4.00\hat{j}$

We need to calculate the acceleration of the wallet

Using formula of acceleration

$a=r\omega^2$

Both the purse and wallet have same angular velocity

$\omega=\omega'$

$\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}$

$\dfrac{a}{r}=\dfrac{a'}{r'}$

$\dfrac{a'}{a}=\dfrac{r'}{r}$

$\dfrac{a'}{a}=\dfrac{3.45}{2.30}$

$\dfrac{a'}{a}=\dfrac{3}{2}$

$a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})$

$a'=3\hat{i}+6\hat{j}$

Hence, The acceleration of the wallet is $3\hat{i}+6\hat{j}$

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3 years ago

A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How

alina1380 [7]

Answer:

(a) The final angular speed is 12.05 rad/s

(b) The time taken to turn 5.5 revolutions is 5.74 s

Explanation:

Given;

number of revolutions, θ = 5.5 revolutions

acceleration of the wheel, α = 20 rpm/s

number of revolutions in radian is given as;

θ = 5.5 x 2π = 34.562 rad

angular acceleration in rad/s² is given as;

$\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2$

(a)

The final angular speed is given as;

$\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 + 2\alpha \theta\\\\\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s$

(b) the time taken to turn 5.5 revolutions is given as

$\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s$

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2 years ago

How is Venus similar to Earth?

hoa [83]

Answer: D

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3 years ago

Read 2 more answers