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Romashka-Z-Leto [24]
3 years ago
6

An aluminum bar has a mass of 9 kg in the air. Calculate its volume. Now imagine that you submerge the aluminum bar in water han

ging it from a rope. Calculate the apparent weight it would have in the water.
Physics
1 answer:
yan [13]3 years ago
7 0

Answer:

55.7 N

Explanation:

The density of aluminum is 2710 kg/m³.  So its volume is:

V = (9 kg) / (2710 kg/m³)

V = 0.00332 m³

The apparent weight is the actual weight minus the buoyant force.

N = mg − B

N = mg − ρVg

N = g (m − ρV)

N = (9.8 m/s²) (9 kg − (1000 kg/m³) (0.0332 m³))

N = 55.7 N

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A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.what a
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Answer:

I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.

IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43

It could have any direction of

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A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f
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Answer:

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Explanation:

Metric unit conversion

100g = 0.1 kg

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As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

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Therefore, as elastic energy is converted to potential energy and work of friction:

E_e = W_f + E_p

kx^2/2 = 7.5 + 1.5 = 9 J

k = 9*2/x^2 = 18/0.05^2 = 7200 N/m

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