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serious [3.7K]
3 years ago
8

A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em

pty, aside from containing air. The bucket is then inverted so that its open end is down and, being careful not to lose any of the air trapped inside, the bucket is lowered below the surface of a fresh-water lake so the water-air interface in the bucket is 20.0 m below the surface of the lake.
To keep the calculations simple, use g = 10 m/s2, atmospheric pressure = 1. 105 Pa, and the density of water as 1000 kg/m3.a. If the temperature is the same at the surface of the lake and at a depth of 20.0 m below the surface, what is the height of the cylinder of air in the bucket when the bucket is at a depth of 20.0 m below the surface of the lake?b. If, instead, the temperature changes from 300 K at the surface of the lake to 275 K at a depth of 20.0 m below the surface of the lake, what is the height of the cylinder of air in the bucket?
Physics
1 answer:
Andrew [12]3 years ago
6 0

Answer:

12 cm

Explanation:

P_1 = Initial pressure = P_a=1\times 10^5\ Pa

P_2 = Final pressure = P_a+\rho_w gh

h = Depth of cylinder = 36 cm

g = Acceleration due to gravity = 10 m/s²

\rho_w = Density of water = 1000 kg/m³

h_1 = Depth of lake = 20 m

From the ideal gas relation we have

P_1V_1=P_2V_2\\\Rightarrow P_a(\pi r^2h)=(P_a+\rho_w gh_1)\pi r^2h'\\\Rightarrow 1\times 10^5\times 36=(1\times 10^5+1000\times 10\times 20)h'\\\Rightarrow h'=\dfrac{1\times 10^5\times 36}{1\times 10^5+1000\times 10\times 20}\\\Rightarrow h'=12\ cm

The height of the cylinder of air in the bucket when the bucket is at the given depth is 12 cm

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A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee
emmainna [20.7K]

a) 0.0028 rad/s

b) 23.68 m/s^2

c) 0 m/s^2

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

\omega = \frac{\theta}{t}

where

\theta is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

\omega = \frac{v}{r} (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

v=29,960 km/h is the linear speed, converted into m/s,

v=8322 m/s

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the  centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

a_r=\omega^2 r

where

\omega is the angular speed

r is the radius of the orbit

For the spaceship in the problem, we have

\omega=0.0028 rad/s is the angular speed

r=2925 km = 2.925\cdot 10^6 m is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

a_t=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in the linear speed

\Delta  t is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

v=8322 m/s

Therefore, its linear speed is not changing, so the change in linear speed is zero:

\Delta v=0

And therefore, the tangential acceleration is zero as well:

a_t=\frac{0}{\Delta t}=0 m/s^2

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3 years ago
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Answer:

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La velocidad correcta de la luz en el vacío es  300.000 km/s .

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Duración de tiempo = 494 segundos, porque cada minuto = 60 segundos

La distancia = (300.000 km/s) x (494 s)

<em>La distancia = 148.200.000 km</em>

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A wire in a circuit carries a current of 0.9 A. Calculate the quantity of
raketka [301]

Answer:

We conclude that the quantity of  the charge that flows through the wire in 50 s will be 45 C.

Explanation:

Given

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  • Time t = 50 s

To determine

We need to find the quantity of  the charge that flows through the wire in 50 s.

Important Tip:

A current of 1 Ampere = 1 Coulomb  of charge flowing in 1 second

Using the formula involving charge and current

I=\frac{Q}{t}

where:

  • I represents the current in amperes (A)
  • Q represents the charge in coulomb (C)
  • t represents the time in seconds (s)

now substituting I = 0.9 and t = 50 in the formula

I=\frac{Q}{t}

0.9\:=\:\frac{Q}{50}

switch sides

\frac{Q}{50}=0.9

Multiply both sides by 50

\frac{50Q}{50}=0.9\cdot \:50

Simplify

Q=45 C

Therefore, we conclude that the quantity of  the charge that flows through the wire in 50 s will be 45 C.

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