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Shtirlitz [24]
3 years ago
12

Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.

Physics
1 answer:
Mandarinka [93]3 years ago
3 0

Answer:

4.9 x 10⁻⁴N

Explanation:

Given parameters:

Mass of the vertical strand of spiderweb  = 5 x 10⁻⁵kg

Unknown:

Tension in the vertical strand  = ?

Solution:

The tension is the vertical strand will be the weight of the strand.

  Weight of strand  = mg

m is the mass

g is the acceleration due to gravity = 9.8m/s²

Tension in the vertical strand  = 5 x 10⁻⁵kg  x 9.8m/s²

Tension in the vertical strand  = 4.9 x 10⁻⁴N

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3 years ago
slab of ice floats on water with a large portion submerged beneath the water surface. The slab is in the shape of a rectangular
n200080 [17]

Answer:

a) \%V = 87.36\,\%, b) x = 1.248\,m, c) F_{B} = 176488.341\,N, d) Six polar bears.

Explanation:

a) The slab of ice is modelled by the Archimedes' Principles and the Newton's Laws, whose equation of equilibrium is:

\Sigma F =\rho_{w}\cdot g \cdot A \cdot x-\rho_{i}\cdot g\cdot V = 0

The height of the ice submerged is:

\rho_{w}\cdot A \cdot x = \rho_{i}\cdot V

x = \frac{\rho_{i}\cdot V}{\rho_{w}\cdot A}

x = \frac{\left(900\,\frac{kg}{m^{3}}\right)\cdot (20\,m^{3})}{\left(1030\,\frac{kg}{m^{3}} \right)\cdot (14\,m^{2})}

x = 1.248\,m

The percentage of the volume of the ice that is submerged is:

\%V = \frac{(1.248\,m)\cdot (14\,m^{2})}{20\,m^{3}} \times 100\,\%

\%V = 87.36\,\%

b) The height of the portion of the ice that is submerged is:

x = 1.248\,m

c) The buoyant force acting on the ice is:

F_{B} = \left(1030\,\frac{kg}{m^{3}} \right)\cdot (1.248\,m)\cdot (14\,m^{2})\cdot \left(9.807\,\frac{m}{s^{2}} \right)

F_{B} = 176488.341\,N

d) The new system is modelled after the Archimedes' Principle and Newton's Laws:

\Sigma F = -n\cdot m_{bear}\cdot g-\rho_{i}\cdot V \cdot g + \rho_{w}\cdot V\cdot g = 0

The number of polar bear is cleared in the equation:

n\cdot m_{bear} = (\rho_{w} - \rho_{i})\cdot V

n = \frac{(\rho_{w}-\rho_{i})\cdot V}{m_{bear}}

n = \frac{\left(1030\,\frac{kg}{m^{3}} - 900\,\frac{kg}{m^{3}} \right)\cdot (20\,m^{3})}{400\,kg}

n = 6.5

The maximum number of polar bears that slab could support is 6.

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