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Semmy [17]
2 years ago
12

What is the maximum mass that can hang without sinking from a 50-cm diameter Styrofoam sphere in water

Physics
1 answer:
ddd [48]2 years ago
8 0

This question is incomplete, the complete question is;

Styrofoam has a density of 32kg/m³.

What is the maximum mass that can hang without sinking from a 50-cm diameter Styrofoam sphere in water

Answer: the required maximum mass is 63.3459 kg

Explanation:

Given that;

diameter d = 50 cm = 0.5 m

radius r = 0.5 / 2 = 0.25 m

Styrofoam density = 32kg/m³

Volume of sphere V = 4/3 πr³.

we substitute

V = 4/3 π(0.25)³

V = 0.06544 m³

Mass of sphere M = Volume × Density

we substitute

Mass of sphere M = 0.06544 m³ × 32kg/m³

Mass of Sphere M = 2.09408 kg

Mass of the water it displace will be;

volume × density

we know that density of water = 1000 kg/m³

so

Mass of the water it displace = 0.06544 m³ × 1000kg/m³

= 65.44 kg

Now the difference between mass of water and mass of the Styrofoam will be the amount of mass that the sphere can support

so

65.44 kg - 2.09408 kg = 63.3459 kg

Therefore, the required maximum mass is 63.3459 kg

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a_{rad}= \frac{v^2}{r}
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8 0
3 years ago
In classical physics, consider a 2 kg block hanging on a spring with a spring constant of 50 N/m. Ignore air resistance. The blo
RUDIKE [14]

Answer:

v = 0

Explanation:

This problem can be solved by taking into account:

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T = \sqrt{\frac{m}{k} }     ( 1 )

- The equation for the velocity of a simple harmonic motion

x = \frac{2\pi }{T}Asin(\frac{2\pi }{T}t)   ( 2 )

where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block

Hence

T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s

and by reeplacing it in ( 2 ):

v = \frac{2\pi }{0.2s}(14cm)sin(\frac{2\pi }{0.2s}(0.9s)) = 140\pi  sin(9\pi ) = 0

In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.

5 0
3 years ago
I run around a circular track, with radius 25m, for 4 and times before stopping. It takes me 16
Sergio039 [100]

Answer:

13m/s. is the answer probably but do u have ms gallup too?

4 0
2 years ago
At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex
telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let F_{e} be the force of gravity exerted by the earth

and let F_{m} be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d,  is given by:

F = \frac{Gm_{1} m_{2} }{d^{2} }

Mass of the earth, m_{e} = 5.97 * 10^{24} kg

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Mass of the satellite, m_{s} = ?

F_{e}  = \frac{G*5.97 * 10^{24} M }{d^{2} }...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

F_{m}  = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }............................(2)

Equating equations (1) and (2)

\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }

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Factorising out 10^{24}

1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0

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d = 346 * 10⁶ m

4 0
3 years ago
A piece of plastic with a mass of 15 g is
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Answer:

The density of plastic is equal to 0.6 g/mL.

Explanation:

Given that,

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So, the density of plastic is equal to 0.6 g/mL.

5 0
2 years ago
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