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nignag [31]
2 years ago
8

The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank

has a density of 1.0 g/mL. How far has the boat sunk into the water?
Physics
1 answer:
Paladinen [302]2 years ago
8 0

When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat

So here weight of the boat = Buoyancy force

let say boat is sunk by distance "h"

now we can say

F_b = \rho * V * g

F_b = 1000*0.10 * 0.08 * h * 9.8

now by above force balance equation we can write

m*g = F_b

0.040 * 9.8 = 1000 * 0.10 * 0.08 * h * 9.8

0.040 = 8h

h = 5 * 10^{-3} m

so boat will sunk by total 5 mm distance

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Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

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We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
1 year ago
A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope
pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

L=\dfrac{\lambda}{4}

\lambda=4\times L

Put the value into the formula

\lambda=4\times31.2\times10^{-2}

\lambda=1.248\ m

We need to calculate the frequency of sound in pipe

Using formula of frequency

f=\dfrac{v}{\lambda}

Put the value into the formula

f=\dfrac{340}{1.248}

f=272.4\ Hz

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

L=\dfrac{l}{3}

L = \dfrac{129}{3}

L=43\ cm

We need to calculate the wavelength of the wire

Using formula of length

L=\dfrac{\lambda}{2}

\lambda=L\times 2

Put the value into the formula

\lambda=43\times2

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We need to calculate the speed of the wave

Using formula of frequency

f=\dfrac{v}{\lambda}

v=f\times\lambda

Put the value into the formula

v=272.4\times86\times10^{-2}

v=234.26\ m/s

Hence, The speed of transverse waves in the wire is 234.26 m/s.

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Answer:

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