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AlexFokin [52]
3 years ago
11

You have a piece of cork with a volume of 2 cm^3 and a density of 210kg/m^3. You hold it under water and release it.

Physics
1 answer:
insens350 [35]3 years ago
7 0

I am sorry if it didn't helped

answers;

Calculate the buoyant force of a piece of cork of 8cm3 that floats in water. Density of cork is 207kg/m3. ?

I need the mass, in order to get the volume to apply t to the Buoyancy formula of: B=(W)object=(m)object(g)

Explanation:

From Archimedes Principle, any object partially or totally submerged in a fluid is buoyed upwards with a force equal to the weight of the displaced fluid.

∴

B

=

ρ

f

l

V

f

l

g

=

1000

k

g

/

m

3

×

8

×

10

−

6

m

3

×

9

,

8

m

/

s

2

=

0

,

0784

N

(assuming the density of water is at standard temperature and pressure, and that the cork is totally submerged as it floats in the water

it's not the answer of your question ⁉️ but it is similar ........

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Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

5 0
4 years ago
John doe gets on the highway in his 1967 Shelby 427 Cobra starting from the dead stop at the bottom of the on ramp of it can be
masha68 [24]

Answer:

It will take john 4.477 seconds

7 0
3 years ago
True or False: There is the same amount of water on earth today as there was when earth was formed.
Nataliya [291]

Answer:

Yes the water amounts are the same.

Explanation:

4 0
3 years ago
Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c
mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

V = \frac{k*q}{r}

where k = \frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}

Replacing by the values in (1) we have:

V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q)  = 0 V

which is equal to the option d).

6 0
4 years ago
If the beam carries 1015 electrons per second and is accelerated by a 350 kV source, find the current and power in the beam.
Y_Kistochka [10]

To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,

I = q*N

Here q is Flow of electric charge in one second and N the number of electron flow per second.

A the same time the power is described as the applied voltage for the current.

P = VI

We know the charge of electron, q = 1.602 * 10^{-19} Coulombs, then the current is

I = (1.602*10^{-19})(10^{15})

I = 0.1602 mA

And the power in the Beam is

P = VI

P = (350*10^3)(0.1602)

P = 0.05607 Watts

3 0
3 years ago
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