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docker41 [41]
3 years ago
8

the hot gas lines need to be insulated? A never B always C if they are exposed to cold temperatures that could cause the refrige

rant to condense in the line D for systems using hfc refrigerants
Physics
1 answer:
muminat3 years ago
8 0
The answer is a because it never needs to be insulated
You might be interested in
our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three
Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is \tau = 35.63\ N \cdot m  

b

The distance of the single for equilibrium to occur is r_s =1.62 \ m

Explanation:

From the question we are told that

     The mass of the left rock is  m_s = 2.25 \ kg

     The mass of the rock on the right m_p = 10.1 kg

    The distance from  fulcrum to the center of the pile of rocks is  r_p = 0.360 \ m

   

Generally the torque produced by the pile of rock is mathematically represented as

           \tau = m_p * g * r_p

Substituting values

         \tau = 10.1 * 9.8  * 0.360                  

          \tau = 35.63\ N \cdot m      

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

   The torque due to the single rock is

           \tau = m_s  * g * r_s

At equilibrium the both torque are equal

            35.63 = m_s * r_s * g

Making r_s the subject of the formula

             r_s = \frac{35.63 }{m_s * g}

Substituting values

            r_s = \frac{35.63 }{2.25 * 9.8}

            r_s =1.62 \ m

6 0
3 years ago
the roque requried to turn the crank on an ice cream maker is 4.50 N.m how much work does it take to turn the crank through 300
Alexus [3.1K]

Answer:

the work required to turn the crank at the given revolutions is 8,483.4 J

Explanation:

Given;

torque required to turn the crank, T = 4.50 N.m

number of revolutions, = 300 turns

The work required to turn the crank is given as;

W = 2πT

W = 2 x 3.142 x 4.5

W = 28.278 J

1 revolution = 28.278 J

300 revlotions = ?

= 300 x 28.278 J

= 8,483.4 J

Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J

4 0
3 years ago
In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg. After the two
IrinaVladis [17]

This question involves the concepts of the law of conservation of momentum.

The magnitude of the final momentum of the eight ball is "0.22 N.s".

According to the law of conservation of momentum:

P_{i1}+P_{i2}=P_{f1}+P_{f2}

where,

P_{i1} = initial momentum of the cue ball = 0.23 N.s

P_{i2} = initial momentum of the eight ball = 0 N.s (since ball is initially at rest)

P_{f1} = final momentum of the cue ball = 0.01 N.s

P_{f2} = final momentum of the eight ball = ?

Therefore,

0.23\ N.s + 0\N.s = 0.01\ N.s+P_{f2}\\\\P_{f2} = 0.22\ N.s

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

3 0
3 years ago
: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma
e-lub [12.9K]

Answer:

x_1 = 3.74m

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

a_s = \dfrac{F}{m}

a_s = \dfrac{8.2}{12}

a_s = 0.68\ m/s^2

F = m a_m

a_m = \dfrac{F}{m}

a_m = \dfrac{8.2}{70}

a_m = 0.12\ m/s^2

x_1+x_2 = 25

\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25

(a_c+a_m)t^2=50

(0.12+0.68)t^2=50

t = \sqrt{\dfrac{50}{0.8}}

t = 7.90 s

x_1 = \dfrac{1}{2}a_ct^2

x_1 = 0.5\times 0.12 \times 7.90^2

x_1 = 3.74m

5 0
3 years ago
A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction bet
Vikki [24]

Answer:

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

<u>a = 2.94 m/s²</u>

3 0
3 years ago
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