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wolverine [178]
3 years ago
8

A devout halloweener not only dressed as an astronaut, but travelled to the moon for the full experience. The astronaut jumps on

the moon with an initial upward velocity of 2.98m/s. If the acceleration of gravity on the moon is -1.62m/s2, what is the highest the astronaut jumps?
Physics
1 answer:
nikdorinn [45]3 years ago
5 0

Answer:

2.78 m

Explanation:

At the peak, the velocity is 0.

Given:

a = -1.6 m/s²

v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

x

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (2.98 m/s)² + 2(-1.6 m/s²) (x - 0 m)

x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

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Explanation:

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A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
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Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

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W = 3.12 J

Hope this helps!

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3 years ago
An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t
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Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

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Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = 1.013 \times 10^{5}+ 100 \times 1000 \times 9.8

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Use the Boyle's law

P V = P' V'

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V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

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