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denis-greek [22]
3 years ago
10

A plumber is going to put two pipes in a wall, one in front and one in back. The pipes will be touching once they are installed.

Both pipes are the same size and have the same number of molecules. The diagram above shows the pipes now, before they have touched. Use the information in the diagram to answer the question.
How does the temperature of the front pipe compare with the temperature of the back pipe before the pipes touch? What will happen after the pipes have been touching for a while?
Physics
1 answer:
beks73 [17]3 years ago
8 0

Answer:

They become the same exact tempature

Explanation:

Since they got connected it should be the same.

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How many times does the world go around
sattari [20]
I'm not sure what you're asking but the earth has the ability to infinitely continue to spin or the earth completes 365.25 rotations during a full cycle. 
6 0
3 years ago
An air craft heads north at 320 km/hr relative to the wind. the wind velocity is 80km/hr from the north. find the relative veloc
Gnoma [55]

Answer:

Relative to the ground, the velocity of the aircraft is 240 km/hr

Explanation:

Relative velocity is different from normal velocity;

When 2 objects are moving in opposite directions towards each other, they will appear to be faster than they actually are;

This is known as the relative velocity;

The information tells us we have the aircraft moving 320 km/hr northwards relative to the wind;

The wind is in the opposite direction at 80 km/hr;

R = relative velocity of the aircraft

v = actual velocity of the aircraft

w = velocity of the wind

R = v + w

Note: if the wind was moving in the same direction, the formula would be R = v - w

320 = v + 80

v = 320 - 80

v = 240

The velocity relative to the ground is simply the actual velocity as the ground doesn't move;

So, relative to the ground, the velocity of the aircraft is simply 240 km/hr

7 0
3 years ago
An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of
GaryK [48]

Explanation:

The given data is as follows.

     Velocity of bullet, c_{p} = 814.8 m/s

    Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

               t = \frac{24.7}{343}      (velocity in air = 343 m/s)

                 = 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

               d_{b} = c_{b} \times t

                          = 814.8 \times 0.072

                          = 58.66

                          = 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0
3 years ago
Free Fall: A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses
Slav-nsk [51]

Answer:

v₀₁= 5.525 m / s

Explanation

Freefall Formulas :

The sign of acceleration due to gravity  (g) is positive if the object is going down and negative if the object is going up.

vf= v₀+gt  

vf²=v₀²+2*g*h

h= v₀t+ (1/2)*g*t²

Where:  

h: hight in meters (m)    

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Kinematics of the rock from the starting point with vo until it reaches its maximum height:

vf₁= v₀₁-gt₁  :vf₁ =0 to maximum height

0= v₀₁-gt₁

v₀₁ = g*t₁

t₁ =v₀₁ / g      Equation (1)

vf₁²= v₀₁²-2*g*h₁   : vf₁ =0 to maximum height

0 = v₀₁²-2*g*h₁

2*g*h₁ = v₀₁²

h₁ = (v₀₁²)/(2g)   Equation (2)

Kinematics of the rock when it falls from the maximum height until it touches the floor

h₂= v₀₂t+ (1/2)*g*t₂²  v₀₂=vf₁ =0

h₂= 0+ (1/2)*g*t₂²

h₂= (1/2)*g*t₂²   Equation (3)

Equation that relates h₁ to h₂

h₂=  h₁ + 56.3  ,  h₁ = (v₀₁²)/(2g)

h₂= (v₀₁²)/(2g) + 56.3  Equation (4)

Equation that relates t₁ to t₂

t₁ + t₂ =4 s

t₂ =4 -t₁

t₂ =4 -(v₀₁/g )

Calculation of v₀₁

We replace equation 4 and equation 5 in equation 3

(v₀₁²)/(2g) + 56.3 = (1/2)*g*(4 -(v₀₁/g ) )²

(v₀₁²)/(2g) + 56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g )+((v₀₁/g )²)

we eliminate (v₀₁²)/(2g) on both sides of the equation

56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g ))

56.3 = 78.4 - 4*v₀₁

4*v₀₁ =78.4-56.3

v₀₁= (78.4-56.3) / ( 4)

v₀₁= 5.525 m / s

7 0
3 years ago
Nodes and antinodes.are part of an _____ wave<br><br> A active <br> B standing
mel-nik [20]
It could be A :) not sure tho
5 0
3 years ago
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