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belka [17]
3 years ago
12

A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the block in water?

Physics
1 answer:
NeX [460]3 years ago
5 0
If one m³ of that material holds 4,000 kg of it,
then 0.09 m³ holds

(0.09) x (4,000) = 360 kg of it

The force of gravity acting on 360 kg of anything
on the Earth's surface is

(mass) x (gravity)

= (360 kg) x (9.8 m/s²) = 3,528
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U. A hockey player takes a slap shot at a puck at rest on
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Answer:

a) 7200 ft/s²

b) 140 ft

c) 3.7 s

Explanation:

(a) Average acceleration is the change in velocity over change in time.

a_avg = Δv / Δt

We need to find what velocity the puck reached after it was hit by the hockey player.

We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s².  Therefore:

v² = v₀² + 2a(x − x₀)

(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)

v₀² = 5200 ft²/s²

v₀ = 20√13 ft/s

So the average acceleration impacted to the puck as it is struck is:

a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)

a_avg = 2000√13 ft/s²

a_avg ≈ 7200 ft/s²

(b) The distance the puck travels before stopping is:

v² = v₀² + 2a(x − x₀)

(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)

x = 140 ft

(c) The time the puck takes to travel 10 ft without friction is:

t = (10 ft) / (20√13 ft/s)

t = (√13)/26 s

The time the puck travels over the rough ice is:

v = at + v₀

(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)

t = √13 s

So the total time is:

t = (√13)/26 s + √13 s

t = (27√13)/26 s

t ≈ 3.7 s

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SVETLANKA909090 [29]

Answer:

reviewing the statement we can say that it is FALSE

Explanation:

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When reviewing the statement we can say that it is FALSE

6 0
2 years ago
If a 4N weight is hung on a spring, and it extends by 0.2m, what is the spring constant (k)?
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Explanation:

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