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3 years ago

7

In an HVAC system, the purpose of pneumatics is to serve as A. a controller B. conditional air flow C. an energy source D. a sou

rce of conditioning

1 answer:

Naya [18.7K]3 years ago

3 0

Answer: <u><em>A</em></u>

Explanation:

A pneumatic control system uses compressed air as a method of control for HVAC systems. ... Each senor responds to changes in temperature, humidity, and static pressure as examples, to provide feedback in a control loop to open or close the actuator to meet the control set point.

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Is the earth's gravitational force on the sun larger than, smaller than, or equal to the sun's gravitational force on the earth?

Leona [35]

Answer:

The earth's gravitational force on the sun is equal to the sun's gravitational force on the earth

Explanation:

Newton's third law (law of action-reaction) states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In other words, when two objects exert a force on each other, then the magnitude of the two forces is the same (while the directions are opposite).

In this problem, we can call the Sun as "object A" and the Earth as "object B". According to Newton's third law, therefore, we can say that the gravitational force that the Earth exerts on the Sun is equal (in magnitude, and opposite in direction) to the gravitational force that the Sun exerts on the Earth.

6 0

3 years ago

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place

anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

4 0

3 years ago

very fine smoke particles are suspended in air. the translational rms speed of a smoke particle is 2.45 10-3 m/s, and the temper

m_a_m_a [10]

The mass of a particle is 2.2x10⁻¹⁵ kg

Consider smoke particles as an ideal gas

The translational RMS speed of the smoke particles is 2.45x10⁻³ m/s.

<em>v= √3kT/m</em>

<em>where k= 1.38x10⁻²³J/K, T is 288K, and m is the mass of the smoke particle</em>

<em>2.45x10⁻³ = √3x1.38x10⁻²³x288/m</em>

<em>m= 2.2x10⁻¹⁵ kg</em>

Therefore, the mass of a particle is 2.2x10⁻¹⁵ kg.

To learn more about the translational root mean square speed of gases, visit brainly.com/question/6853705

#SPJ4

5 0

1 year ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

3 years ago

How do smaller summer ice flores affect polar bears physically?

rosijanka [135]

Answer:

Bears lose necessary body fat because they must swim longer in open ocean water.

Explanation:

Polar bears, like their cousins, the grizzly bears, must build up a store of body fat that they can live off of during the winter months. This is especially true of mother bears who give birth to cubs during this time and need enough nourishment to feed the cubs.

(can i have brainliest please?)

8 0

3 years ago