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3 years ago

A student increases the temperature of a 300 cm^3 balloon from 30c to 60c. what will the new volume of the ballon be

2 answers:

5 0

<span>This problem uses the ideal gas law, PV = nRT. T for temperature must be in Kelvin, so we add 273.15 to the Celsius temperatures. We are then changing T from 303.15 K to 333.15K. The ratio of the temperatures 333.15/303.15 ~= 1.1. So holding pressure constant, the new volume of the balloon will be approximately 10% larger, or 330cm^3.</span>

Mama L [17]3 years ago

4 0

From Charles law the volume of a fixed mass of gas is directly proportional to the absolute temperature when pressure is kept constant.
Volume α temperature, mathematically; k = V/T
Therefore, comparing two gases V1/T1 = V2/T2
V1 =300 cm³ T1= 30°C (30 +273= 303K) V2= ? T2= 60°C (60 +273= 333K)
Thus, 300/303 = V2/333
V2 = (300 × 333)/ 303
= 329.703 cm³
Hence the new volume will be 329.703 cm³

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16. A student uses 110 N of force to push a box

Lubov Fominskaja [6]

Answer:

792J

Explanation:

Use the formula Force × Distance

With 110 newtons being the force and 7.2 meters being the distance

Which gives 792 and work us measured in joules therefore its joules

5 0

3 years ago

iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms

saw5 [17]

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

$\frac{2}{a^{3} } = \frac{N_{A} }{V_{molar} }$ where $V_{molar} = \frac{A}{\rho }$ an and A is the atomic mass of iron.

Therefore:

$\frac{2}{a^{3} } = \frac{N_{A} * p }{A}$

This implies that:

$A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} } }$

= $\frac{4}{\sqrt{3} }r$

Assuming that there is no phase change gives:

$\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3} }$

= 8.60 g/m³

3 0

3 years ago

Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0

3 years ago

Read 2 more answers

A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o

aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

$E_r = \frac{1}{2} mv^2$