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sweet-ann [11.9K]
3 years ago
7

A hot air balloon is filled with 1.45 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 109 ∘ C . Wha

t is the volume of the air in the balloon after it is heated?
Physics
1 answer:
just olya [345]3 years ago
6 0

Answer:

<em>The volume of air after the balloon is heated = 1.95 × 10⁶ L</em>

Explanation:

Charles Law: Charles' law states that the volume of a given mass of gas is directly proportional to the temperature in Kelvin, provided that the pressure remains constant.

It can be expressed mathematically as,

V₁/T₁ = V₂/T₂

Making V₂ The subject of the equation

V₂ = (V₁/T₁)T₂.................... Equation 1

Where V₁ = Initial Volume, T₁ = Initial Temperature, V₂ = Final Volume, T₂ = final Temperature

<em>Given: V₁ = 1.45 × 10⁶ L, T₁ = 11 °C = (11 + 273) K = 284 K, T₂ = 109 °C = (109 + 273) = 382 K.</em>

<em>Substituting these values into equation 1 above,</em>

<em>V₂ = (1.45×10⁶)382/284</em>

<em>V₂ = 1.95 × 10⁶ L</em>

<em>Therefore the volume of air after the balloon is heated = 1.95 × 10⁶ L</em>

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
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Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

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v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

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