Physics, 09.12.2020 21:50 emily41950

A 4 m^3 metal block is suspended using a cable in salt water with density 1030 kg/m^3. Tension on the cable is 42600N. What is the weight of the block. What is the density of the block?

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Answer from: nancyy85

The density of the block is 1113.216 kilograms per cubic meter.

The weight of the block is 43669.237 newtons.

Explanation:

According to the Archimedes' Principle, the drag force experimented by the metal block is equal to the weight of the volume of water displace by the block. Besides, the block has a weight that cannot be neglected and experiments a tension from the cable. Given that the metal block is suspended, then we could consider that block is at rest.

From Newton's Laws of Motion we obtain the following equation of equilibrium:

$\Sigma F = T-\rho_{m}\cdot V_{m}\cdot g + \rho_{w}\cdot V_{m}\cdot g = 0$ (1)

Where:

- Tension on the cable, measured in newtons.

$\rho_{w}$ , $\rho_{m}$ - Densities of salt water and the metal block, measured in kilograms per cubic meter.

$V_{m}$ - Volume of the metal block, measured in cubic meters.

- Gravitational acceleration, measured in meters per square second.

If we know that $T = 42600\,N$ , $\rho_{w} = 1030\,\frac{kg}{m^{3}}$ , $V_{m} = 4\,m^{3}$ , $g = 9.807\,\frac{m}{s^{2}}$ , then the density of the block is:

$T+\rho_{w}\cdot V_{m}\cdot g = \rho_{m}\cdot V_{m}\cdot g$

$\rho_{m} = \frac{T+\rho_{w}\cdot V_{m}\cdot g}{V_{m}\cdot g}$

$\rho_{m} = \frac{42600\,N+\left(1030\,\frac{kg}{m^{3}} \right)\cdot (4\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(4\,m^{3})\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\rho_{m} = 1113.216\,\frac{kg}{m^{3}}$

The density of the block is 1113.216 kilograms per cubic meter.

Lastly, the weight of the block (), measured in newtons:

$W = \rho_{m}\cdot V_{m}\cdot g$ (2)

$W = \left(1113.216\,\frac{kg}{m^{3}}\right)\cdot (4\,m^{3}) \cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 43669.237\,N$

The weight of the block is 43669.237 newtons.

Answer from: Quest

The answer is a 0.64 volts

Answer from: Quest

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.

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A 4 m^3 metal block is suspended using a cable in salt water with density 1030 kg/m^3. Tension on th...

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