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3 years ago

The distance between two stations is 1995 Km. How much time will it take to cover the distance at an average speed of 19KM/hour

Physics

1 answer:

Flura [38]3 years ago

5 0

105 hours or 4.375 days

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What is the potential energy of a 2kg plant that is on a windowsill 1.3 m high?

klio [65]

Answer:

25.48

Explanation:

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2 years ago

Galileo started modern science because he started using ? Rather than speculation

Eva8 [605]

By using the telescope

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3 years ago

An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm

horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

$Q = \frac{\Delta P \pi r^4}{8\eta L}$

now we know that

$Q = 0.06 \times 10^{-6} m^3/s$

radius = 0.2 mm

Length = 6.32 cm

$\eta = 1\times 10^{-3} Pa s$

now we have

$6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}$

$3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}$

$\Delta P = 6028 Pa$

now we have

$P - 1500 = 6028 Pa$

$P = 7528 Pa$

8 0

3 years ago

The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,

Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0

3 years ago

The acceleration due to gravity on the Moon's surface is

Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

$g'=\frac{g}{6}$

We know that weight of an object on Earth is,

$w = m\times g$

$m = \frac{w}{g}$

Similarly, weight on Moon will be

$w' = m\times g'$

$w' = \frac{w}{g}\times\frac{g}{6}$

$w' = \frac{300}{6}$

$w' = 50$

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0

3 years ago