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4 years ago

15

I AN TIMED, HURRY!!

2 answers:

kramer4 years ago

6 0

Answer:

C)0.59 N

Explanation:

By the Achemides principle we know that the buoyancy force on an object which is immersed in a incompressible fluid at rest is equal to the weight of the fluid displaced. So all we have to do is to find the mass of 60ml.

Density = mass / volume

Mass = density × volume

= 1 g/cm³× 60 cm³

{ as 1ml=1cm³}

= 60 g

So force equals to,

Force = weight = mass × gravitational acceleration

= 0.06×9.8 = 0.59 N

zvonat [6]4 years ago

3 0

Answer:

.59

Explanation:

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4 years ago

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<h2>•[ ANSWER ]•</h2>

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8 0

3 years ago

For a body falling freely from rest (disregarding air resistance), the distance the body falls varies directly as the square o

My name is Ann [436]

Answer:

The distance is 199.98 ft.

Explanation:

Given that,

Distance = 512 feet

Time = 88 sec

The distance the body falls varies directly as the square of the time.

$s\propto t^2$

$s=kt^2$

We need to calculate the value of constant

Put the value into the formula

$512=k\times(88)^2$

$k=\dfrac{512}{(88)^2}$

$k=0.06611$

We need to calculate the distance

Using formula of distance again

$s=kt^2$

Put the value into the formula

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4 0

3 years ago

A solid sphere of radius R has a uniform charge density and total charge Q. Find the total energy of the sphere U in terms of

Lana71 [14]

Answer:

$\mathbf{U = \dfrac{3 k_c Q^2_{total}}{5R}}$

Explanation:

From the information given;

The surface area of a sphere = $4 \pi r ^2$

If the sphere is from the collection of spherical shells of infinitesimal thickness = dr

Then,

the volume of the thickness and the sphere is;

V = $4 \pi r ^2 \ dr$

Using Gauss Law

$V(r) = \dfrac{k_cq(r)}{r}$

here,

q(r) =charge built up contained in radius r

since we are talking about collections of spherical shells, to work required for the next spherical shell r +dr is

$-dW= dU = V(r) dq = \dfrac{k_c \ q(r)}{r} \ dq$

where;

$q (r) = \dfrac{4}{3} \pi r^3 \rho$

dq which is the charge contained in the next shell of charge

here dq = volume of the shell multiply by the density

$dq = 4 \pi r^2 \ dr \ \rho$

equating it all together

$dU = \dfrac{k_c \frac{4}{3} \pi r^3 \rho}{r} 4 \pi r^2 \ dr \ \rho = \dfrac{16 \pi^2 \ k_c \ \rho^2}{3} \ r^4 \ dr$

Integration the work required from the initial radius r to the final radius R, we get;

$U = \int^R_0 \ dU$

$U = \int^R_0 \dfrac{16 \pi^2 \ k_c \rho^2}{3} r^4 \ dr$

$U = \int^R_0 \dfrac{16 \pi^2 \ k_c \rho^2}{3} [\dfrac{r^5}{5}]^R_0$

$U = \dfrac{16 \pi^2 k_c \rho^2}{15} \ R^5$

Recall that:

the total charge on a sphere, i.e $Q_{total} = \dfrac{4}{3} \pi R^3 \rho$

Then :

$\mathbf{U = \dfrac{3 k_c Q^2_{total}}{5R}}$

6 0

4 years ago