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3 years ago

Air temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?

1 answer:

3 0

The approximate speed of sound in dry (0% humidity) air, in meters per second, at temperatures near 0 °C, can be calculated from

$c_{air} = (331.3+0.606 \upsilon)$

Here

$\upsilon =$ Temperature in Celsius

Replacing with our values we have that

$\upsilon=58\° C$

$c_{air} = (331.3+0.606*58)$

$c_{air} = 366.1m/s$

Therefore the speed of sound in air at that temperature is 366.1m/s

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It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy whe

boyakko [2]

Answer:

10.4L of water is expended when 1L of crude oil is burned.

Explanation:

This problem requires us to calculate the volume of water that must be expended to absorb the amount of energy released from the burning of 1.00L of crude oil.

In going from water at 18.5°C to steam at 285°C, the water undergoes 3 stages:

Absorption of heat from 18.5°C to 100°C >> phase change to vapour at 100°C >> heating from 100°C to steam at 285°C

Given Cw = 4186 J/kg°C and Cs = 2020 J/kg°C. The latent heat of of vaporization of water Lv = 2.256 ×10⁶ J/kg

The total heat absorbed in the process per kilogram

H = Cw × (100 – 18.5) + Lv + Cs × (285 – 100)

H = 4186 × 81.5 + 2. 256 ×10⁶ + 2020× 185

= 2.696 ×10⁶ J/kg

The amount of heat in J/L of water needed can be calculated as follows:

Density of water = 1000kg/m³ =

1000 kg/m³ × 1m³/1000L = 1kg/L (Basically conversion of density in kg/m³ to kg/L)

Let the volume of water needed be V litres.

Then the mass of water that must be expended = Density × Volume

= 1kg/L × V L = Vkg

The heat that would be absorbed by the water when 1L of crude oil is burned is V×H

= 2.696×10⁶ × V

This is also equal to 2.80×10⁷ J of energy (given).

So,

2.696×10⁶V = 2.8×10⁷

V = (2.80×10⁷)/(2.696×10⁶) = 10.4L of water.

3 0

3 years ago

Read 2 more answers

An object moves uniformly around a circular path of radius 19.0 cm, making one complete revolution every 2.40 s.

klio [65]

Answer:

v = 0.5 m/s

f = 0.42 Hz

ω = 2.6 rad/sec

Explanation:

By definition, the translational speed is the rate of change of the position with respect to time.
The change in position along a complete revolution is just the following:
Δs = 2*π*r = 2*π*0.19 m = 1.19 m
The time needed to complete a revolution is 2.4 s, so the translational speed can be written as follows:

$v =\frac{\Delta s}{\Delta t} = \frac{1.19m}{2.4s} = 0.5 m/s (1)$

The frequency in Hz is just the inverse of the time needed to complete a revolution (known as the period T), as follows:
f = 1/T = 1/2.4s = 0.42 Hz (2)
Finally, the angular speed is the rate of change of the angle rotated with respect to time, as follows:

$\omega = \frac{\Delta\theta}{\Delta t} = \frac{2*\pi}{2.4s} = 2.6 rad/sec (3)$

5 0

3 years ago

The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the

ValentinkaMS [17]

Answer:

True

Explanation:

The angular momentum around the center of the planet and the total mechanical energy will be preserved irrespective of whether the object moves from large R to small R. But on the other hand the kinetic energy of the planet will not be conserved because it can change from kinetic energy to potential energy.

Therefore the given statement is True.

5 0

3 years ago

If an electron is lost or gained, we no longer call it an atom. What is the name when this happens?

gtnhenbr [62]

It is called an ion (positive or negative ion)

3 0

3 years ago

You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought g

bezimeni [28]

Answer:

It's only 1.11 m/s2 weaker at 400 km above surface of Earth

Explanation:

Let Earth radius be 6371 km, or 6371000 m. At 400km above the Earth surface would be 6371 + 400 = 6771 km, or 6771000 m

We can use Newton's gravitational law to calculate difference in gravitational acceleration between point A (Earth surface) and point B (400km above Earth surface):

$g = G\frac{M}{r^2}$

where G is the gravitational constant, M is the mass of Earth and r is the distance form the center of Earth to the object

$\frac{g_B}{g_A} = \frac{GM/r^2_B}{GM/r^2_A}$

$\frac{g_B}{g_A} = \left(\frac{r_A}{r_B}\right)^2$

$\frac{g_B}{g_A} = \left(\frac{6371000}{6771000}\right)^2$

$\frac{g_B}{g_A} = 0.94^2 = 0.885$

$g_B = 0.885 g_A$

So the gravitational acceleration at 400km above surface is only 0.885 the gravitational energy at the surface, or 0.885*9.81 = 8.7 m/s2, a difference of (9.81 - 8.7) = 1.11 m/s2.

6 0

3 years ago