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diamong [38]
3 years ago
7

The gas in a balloon has T=280K and V=0.0279m^3. If the temperature increases to 320K at constant pressure, what is the new volu

me of the balloon? (Hint: n and P are constant) (Unit= m^3)
Physics
1 answer:
photoshop1234 [79]3 years ago
7 0

Answer:

\boxed{ V_{2}= 0.03189 m^3}

Explanation:

According to Charles Law

=> \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Where V_{1} = 0.0279 m³, T_{1} = 280 K and T_{2} = 320 K

=> \frac{0.0279}{280} = \frac{V_{2}}{320}

=> V_{2} = 0.03189 m³

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a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
3 years ago
Hi i do not have a question i just want to show u my cute doggo :)) ik everyone on here is probably stressed so hope this makes
Arisa [49]

Answer:

wooowwwwww

Explanation:

it's so so cuteeeeeeee

I love dogs tho:))

5 0
3 years ago
Read 2 more answers
Which would have the longer orbital period: a moon 1 million km from the center of Jupiter, or a moon 1 million km from the cent
Harman [31]

Answer:

earth

Explanation:

The formula for the orbital period of the moon is given by

T = 2\pi \sqrt{\frac{r}{g}}

As the time period is inversely proportional to the square root of the acceleration due to gravity of the planet.

As the value of acceleration due to gravity on Jupiter is more than the earth, so the period of moon around the earth is large as compared to the period of the moon around the Jupiter when the distance is same.

5 0
3 years ago
A system that can be affected by the outside environment, by an exchange of matter or energy is ______.
larisa86 [58]

\sf{Answer:}

A system that can be affected by the outside environment, by an exchange of matter or energy is an open physical system .

4 0
2 years ago
Two long, straight wires are parallel and 26 cm apart.
mezya [45]

Answer: 2.49×10^-3 N/m

Explanation: The force per unit length that two wires exerts on each other is defined by the formula below

F/L = (u×i1×i2) / (2πr)

Where F/L = force per meter

u = permeability of free space = 1.256×10^-6 mkg/s^2A^2

i1 = current on first wire = 57A

i2 = current on second wire = 57 A

r = distance between both wires = 26cm = 0.26m

By substituting the parameters, we have that

Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26

= 4080.744×10^-6/ 1.634

= 4.080×10^-3 / 1.634

= 2.49×10^-3 N/m

5 0
3 years ago
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