Ask question

Ask question

All categories

2 years ago

9

The anemometer is spinning and the pressure us high

1 answer:

skelet666 [1.2K]2 years ago

3 0

Answer:

what help you need?????????

You might be interested in

Adding heat to a liquid causes which of the following physical changes?

Nadya [2.5K]

Decreases density because gases have less density than liquids

4 0

3 years ago

Read 2 more answers

The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e

Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed ( $v$ ), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

$v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi$ (1)

Where:

$\Delta t$ - Rotation time, measured in seconds.

$R$ - Radius of the Earth, measured in meters.

$\phi$ - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that $\Delta t = 86160\,s$ , $R = 6.371\times 10^{6}\,m$ and $\phi = 37.6819^{\circ}$ , then the tangential speed at Livermore is:

$v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}$

$v\approx 284.001\,\frac{m}{s}$

The tangential speed at Livermore is approximately 284.001 meters per second.

4 0

3 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

4 0

3 years ago

A car jack with a mechanical advantage of 6 needs to produce an output force of 600 N to raise a car. What input force is requir

Mkey [24]

B.) 100 N is the CORRECT answer

Please mark as brainliest! :)

6 0

3 years ago

Read 2 more answers

A gas in a sealed container has a pressure of 50 kPa at 27°C. What will the pressure of the gas be if the temperature rises to 8

alexgriva [62]

Answer:

the final pressure of the gas is 60 kPa.

Explanation:

Given;

initial pressure of the gas, P₁ = 50 kPa = 50,000 Pa

initial temperature of the gas, T₁ = 27⁰ C = 27 + 273 = 300 k

final temperature of the gas, T₂ = 87⁰ C = 87 + 273 = 360 K

Let the final pressure of the gas = P₂

Apply pressure law;

$\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} = \frac{50,000 \times 360}{300} = 60,000 \ Pa = 60 \ kPa$

Therefore, the final pressure of the gas is 60 kPa.

4 0

3 years ago