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dusya [7]
2 years ago
14

A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin

al pressure? 1. 0.75 atm 2. 3.0 atm 3. 0.063 atm 4. 0.38 atm 5. 0.12 atm
Physics
1 answer:
ollegr [7]2 years ago
8 0
Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
pV=k
which can also be rewritten as
p_1 V_1 = p_2 V_2 (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, p_1 = 1.5 atm, V_1 =6.5 L and V_2 =13.0 L, so the final pressure of the gas can be found by re-arranging eq.(1):
p_2 = p_1  \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm

Therefore the correct answer is
<span>1. 0.75 atm</span> 
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The short vertical parts adjacent to it also reach into the magnetic field and should experience forces. why can we neglect them
hammer [34]

It is not only the horizontal part of the loop that dips into the magnetic field. We can neglect or disregard the horizontal parts of the loop that hollows into the magnetic fields since only the parts perpendicular to the magnetic field complement to it.

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4 0
2 years ago
Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to
german

Answer: v= 1.23 m/s θ = 75.3º

Explanation:

First of all, we define the direction in which both fruits are tossed as the x axis, so all initial momenta have horizontal components only.

Now, if no external forces act during collision (due to the infinitesimal time during which collision takes place) momentum must be conserved.

As momentum is a vector, both components must be conserved, so we can write the following equations:

p₁ₓ = p₂ₓ ⇒ -m₁ . vi₁ +m₂. vi₂ = m₁ . vf₁ . cos θ  (1)

p₁y = p₂y ⇒ 0 =m₂ . vf₂ - m₁. vf₁. sin θ (2)

Replacing by the values of m1, m₂, vi₁, vi₂, and vf₂, we can calculate the value of the angle θ, that the apple forms with the horizontal, as follows:

(1) -0.13 Kg. 1.05 m/s + 0.15 Kg. 1.18 m/s = 0.13. vf . cos θ  

(2) 0.15 Kg. 1.03 m/s = 0.13 vf. sin θ

sin θ / cos θ = 3.82 ⇒ tg θ = 3.82 ⇒ θ = arc tg (3.82) = 75.3º

Replacing this value of θ in (2), we get:

0.15 kg. 1.03 m/s = 0.13 vf . sin 75.3º = 0.13 . vf . 0.967

Solving for vf, we get:

vf = 0.15 kg. 1.03 m/s / 0.13. 0.967 = 1.23 m/s

5 0
2 years ago
What do fuel cells, batteries and, solar cells have in common A.the all produce static electricity B. they are all sources of di
Vaselesa [24]
The correct answer is B :)
3 0
2 years ago
Read 2 more answers
The mass of the man is 86 kg. What is the force of friction that slowed him down? Use the equation for Newton’s second law, F =
bazaltina [42]

Answer:

  1. Mass = 86 kg acceleration due to gravity = 10m/s

Explanation:

So force =m*a

8 0
2 years ago
Read 2 more answers
A bag containing originally 60 kg of flour is lifted through a vertical distance of 9 m. While it is being lifted, flour is leak
Art [367]

Answer:

The total work done is 5997.6 J

Solution:

As per the question:

Mass of the bag, m = 60 kg

Vertical distance, h = 9 m

Mass lost, m' = 12 kg

To calculate the amount of work done:

Lost mass is proportional to the square root of the distance covered while lifting:

m' ∝ \sqrt{h}

m' = K\sqrt{9}

where

K = proportionality constant

12 = 3K

K = 4

Mass of the floor containing bag at a height h:

m(h) = 60 - k\sqrt{h}

Work done is given by:

W = \int_{0}^{h}m(h)gdh

W = \int_{0}^{9}(60 - k\sqrt{h})gdh

W = g([60h]_{0}^{9} + 4\times \frac{2}{3}[h^{\frac{3}{2}}]_{0}^{9})

W = 9.8\times ([60\times 9 - 0] + \frac{8}{3}[9^{\frac{3}{2}} - 0^{\frac{3}{2}}])

W = 9.8\times (540 + \frac{8}{3}\times 27) = 5997.6\ J = 5.9976\ kJ

3 0
2 years ago
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