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kykrilka [37]
3 years ago
6

The air in a tire pump has a volume of 1.5 L at a temperature of 5 ℃. If the temperature is increased to 25 ℃ and the pressure r

emains constant, what is the new volume of the tire pump
Physics
1 answer:
tankabanditka [31]3 years ago
5 0

Answer:

Explanation:

5 C = 278 K

25 C = 298 K

V1 / T1 = V2 / T2

1.5L / 278 K = V2 / 298 K

V2 = (1.5L * 298) / 278

V2 = 1.61 L

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How long did it take the flag to rotate once in a full circle
Tanzania [10]

Answer:

360 degrees is one full rotation starting at zero

4 0
2 years ago
If a 10kg block is at rest on a table and a 1200N force is applied in the eastward direction for 10 seconds, what is the acceler
gavmur [86]

Answer:

a = 120 m/s²

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a Formula (1)

Known data

Where:

∑Fₓ: Algebraic sum of forces in the x direction

F: Force in Newtons (N)

m: mass (kg)

a: acceleration of the block (m/s²)

F = 1200N

m = 10 kg

Problem development

We replace the known data in formula (1)

1200 = 10*a

a = 1200/10

a = 120 m/s²

6 0
3 years ago
A particle executes simple harmonic motion with an amplitude of 2.18 cm.
Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is  :

                          v (max) = Aω

Speed v at any displacement y is given by  

v^{2} = w^{2} (A^{2} - y^{2})   ........................................................  i

And,

               v = \frac{1}{2} v (max)  

          or,  2 × v = Aω     ....................................................   ii

Eliminating  ω from equations i and ii,

                       \frac{1}{4} A^{2}  w^{2}  =  w^{2}  ( A^{2}  - y^{2})

                     or, y^{2} =  (\frac{3}{4}) A^{2}  =(\frac{3}{4}) 2.18^{2}

                    or,  y =  3.56 cm.

3 0
3 years ago
​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th
vlada-n [284]

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0
3 years ago
A scientist wants to model an internal organ with connective tissue as a mass on a spring. The mass of the organ is 2.0 kg, and
vichka [17]

Answer:

Spring constant, k = 5483.11 N/m

Explanation:

It is given that,

Mass of the organ, m = 2 kg

The natural period of oscillation is, T = 0.12 s

Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

k=\dfrac{4\pi^2 m}{T^2}

k=\dfrac{4\pi^2 \times 2\ kg}{(0.12\ s)^2}

k = 5483.11 N/m

So, the  spring constant for the spring in the scientist's model is 5483.11 N/m.

5 0
4 years ago
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