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

Iwould choose the same momentum because if they throw a baseball very light and not hard at all then doing it with a bowling ball won’t hurt that worse

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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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...