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Natasha2012 [34]

3 years ago

13

An air-conditioning system's automatic controller might directly control the A. conversion of pneumatic energy to hydraulic sign

al pressure. B. thermostat's setpoint temperature C. motor of a large condenser fan D. sensitivity of the ladder logic

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

Answer:

A i think so

Explanation:

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What is the magnitude of the resultant vector? Round your answer to the nearest tenth

Harlamova29_29 [7]

Answer:13.9 on edgy I just did it

Explanation:

7 0

3 years ago

A cylinder of gas at room temperature has a pressure . To p_{1} what temperature in degrees Celsius would the temperature have t

grandymaker [24]

In order to calculate the temperature, we need to know that temperature and pressure are directly proportional, that is, if the pressure increases, the temperature (in Kelvin) also increases in the same proportion.

So, first let's convert the temperature from Celsius to Kelvin, by adding 273 units:

$\begin{gathered} K=C+273 \\ K=20+273 \\ K=293 \end{gathered}$

Then, let's calculate the proportion:

$\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ \frac{p_1}{293}=\frac{1.5p_1}{T_2} \\ \frac{1}{293}=\frac{1.5}{T_2} \\ T_2=1.5\cdot293 \\ T_2=439.5\text{ K} \end{gathered}$

Now, converting back to Celsius, we have:

$\begin{gathered} C=K-273 \\ C=439.5-273 \\ C=166.5\text{ \degree{}C} \end{gathered}$

So the temperature would be 166.5 °C.

6 0

1 year ago

In measuring the width of a hair sample, a light of wavelength 500 nm is used. The hair sample is 40 um in radius. With the scre

sergejj [24]

Answer:

The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

Explanation:

Given that,

Wave length = 500 nm

Radius $d= 40\ \mu m$

Distance from the hair sample D= 6 m

We need to calculate the distance of the second dark band away from the central bright spot be located

$\sin\theta=\dfrac{y}{D}$

$\sin\theta=\dfrac{y}{6}$

Using formula for dark fringe

$(n-\dfrac{1}{2})\lambda=2d\sin\theta$

Put the value into the formula

$(2-\dfrac{1}{2})\times500\times10^{-9}=2\times40\times10^{-6}\times\dfrac{y}{6}$

$y=\dfrac{(2-\dfrac{1}{2})\times500\times10^{-9}\times6}{2\times40\times10^{-6}}$

$y=0.05625\ m$

$y=5.625\times10^{-2}\ m$

Hence, The distance of the second dark band away from the central bright spot be located is $5.625\times10^{-2}\ m$

6 0

3 years ago

How many significant figures are in the number 17.95?

ANTONII [103]

Answer:

4

Explanation:

All of them are significant figures

6 0

4 years ago

An object initially at rest accelerates at 5 meters per second2 until it attains a speed of 30 meters per second. What distance

Tcecarenko [31]

Answer: 90m

Explanation:

Use Equation for distance:

S=a*t²/2

use eqation for acceleraton a=(Vf-Vs)/t

Vs- starting speed

Vf - final speed

a-accelaration

---------------------------------

a=5m/s²

Vs=0m/s

Vf=30m/s

use

a=(Vf-Vs)/t

to find time

t=(Vf-Vs)/a

t=30m/s/5m/s²

t=6s

Now calculate distance that object travel using:

S=(a*t²)72

s=(5m/s²*(6s)²)/2

S=90m

4 0

3 years ago

Read 2 more answers