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3 years ago

15

I have a question, it concerns hydrostatic buoyancy and Archimedes' law.

1 answer:

saw5 [17]3 years ago

4 0

Answer: the lvl wud remain the same

Explanation: as per Archimedes Principle, the weight of the water displaced by the object is equal to the weight of the object. When the ship initially went into the pool, it wud hv displaced some water. When the anchor is dropped, the level does not change coz the anchor was already in the ship and no extra weight has been added, so the weight of the anchor has already been accounted for in the first place when the ship was first placed in the pool

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3 years ago

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Una bala calibre se deja caer desde el piso 84 (a una altura de 250 metros) La bala tiene una masa de 0,008 kg. ¿A qué velocidad

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3 years ago

A skier is accelerating down a 30.0-degree hill at 3.80 m/s^2.

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Answer:

ax = -3.29[m/s²]

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3 years ago

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3 years ago

Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.

givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/ $s^{2}$ , t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

= 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

= 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + $\frac{1}{2}$ a $t^{2}$

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = $\frac{1}{2}$ a $t^{2}$

= $\frac{1}{2}$ x 14 x $(510)^{2}$

= 7 x 260100

= 1820700 m

The distance that the shuttle has traveled during the given time is 1820.7 Km.

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2 years ago