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Vsevolod [243]
3 years ago
14

A balloon of hydrogen is put into to pressure chamber. The initial pressure and volume of hydrogen is 1 atm and .5 cm3. The pres

sure of the chamber is increased to 2 atm. What is the new volume? Use the formula for Boyle's Law: P1V1 = P2V2
Physics
1 answer:
Sedbober [7]3 years ago
3 0

Answer:

2.5 cm³

Explanation:

Use Boyle's Law

1×5=2×V

5=2V

V=5/2=2.5 cm³

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An open beaker of pure water has a water potential of.
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Answer: Having Pure Water Is Zero.

Explanation: ...

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n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a
Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

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3 years ago
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A light horizontal spring has a spring constant of 167 N/m. A 2.77 kg block is pressed against one end of the spring, compressin
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Answer:

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5 0
2 years ago
9)A 64 kg parent and a 16 kg child meet at the center of an ice rink. They place their hands together and push. (A) Is the
Colt1911 [192]

Answer:

(A) <u>The same</u>

(B) <u>More</u>

(C) The magnitude of the parent's acceleration is 0.625 m/s²

Explanation:

The given parameters are;

The mass of the parent, m₁ = 64 kg

The mass of the child, m₂ = 16 kg

(A) By Newton's third law of motion, action and reaction are equal and opposite

Therefore, the action of the parent on the child is equal to the reaction of the child on the parent and vice versa

Therefore, the force experienced by the child is <u>the same</u> as the force experienced by the parent

(B) Newton's second law states that an objects acceleration is directly proportional to the applied force and inversely proportional to the mass of the object

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Let F₁ represent the force experienced by the parent, let F₂ represent the force experienced by the child and let a₁ represent the magnitude of the parent's acceleration

By Newton's third law, we have;

F₁ = F₂

Force, F = Mass, m × Acceleration, a

We can write, F = m × a

Therefore;

F₁ = m₁ × a₁ and F₂ = m₂ × a₂

∴ F₁ = F₂ gives;

m₁ × a₁ = m₂ × a₂

a₁ = (m₂ × a₂)/m₁ = (16 × 2.5)/64 = 0.625

∴ The magnitude of the parent's acceleration = a₁ = 0.625 m/s²

5 0
2 years ago
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