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harina [27]
3 years ago
12

2. A construction consultant may be responsible for

Physics
1 answer:
Brrunno [24]3 years ago
3 0

This is a tricky one but on my part I'd have to say depending on the contract A,B,C.

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Describe how cell membranes are selectively permeable
attashe74 [19]
<span>Cell membranes are selectively permeable because it allows some things to enter or leave the cell while keeping other things outside or inside the cell.</span>
5 0
3 years ago
What is first class lever ? can you pease say where is does the fulcrum ,load and effort​
valkas [14]

Answer:

In first class lever the fulcrum lies between the effort and the load the effort lies on the left side and load lies on the right side .

Explanation:

hope this helps

7 0
3 years ago
Boston Red Sox pitcher Roger Clemens could routinely throw a fastball at a horizontal
azamat

Answer: 0.145 seconds

Explanation:

Given that Roger Clemens could routinely throw a fastball at a horizontal speed of 119.7 m/s. How long did the ball take to reach home plate 17.3 m away

Since the speed is horizontal

Using the formula for speed which is

Speed = distance/time

Where

Speed = 119.7 m/s

Distance covered = 17.3 m

Time is what we are looking for

Substitute all the parameters into the formula

119.7 = 17.3/ time

Make time the subject of formula

Time = 17.3 / 119.7

Time = 0.145 seconds.

Therefore, it will take 0.145 seconds to reach the home plates

6 0
3 years ago
5. A massless string passes over a frictionless pulley and carries
devlian [24]

Answer:

2m₁m₃g / (m₁ + m₂ + m₃)

Explanation:

I assume the figure is the one included in my answer.

Draw a free body diagram for each mass.

m₁ has a force T₁ up and m₁g down.

m₂ has a force T₁ up, T₂ down, and m₂g down.

m₃ has a force T₂ up and m₃g down.

Assume that m₁ accelerates up and m₂ and m₃ accelerate down.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₁ − T₂ − m₂g = m₂(-a)

T₁ − T₂ − m₂g = -m₂a

(m₁g + m₁a) − T₂ − m₂g = -m₂a

m₁g + m₁a + m₂a − m₂g = T₂

(m₁ − m₂)g + (m₁ + m₂)a = T₂

Sum of the forces on m₃:

∑F = ma

T₂ − m₃g = m₃(-a)

T₂ − m₃g = -m₃a

a = g − (T₂ / m₃)

Substitute:

(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂

(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂

(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂

m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂

2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂

T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)

8 0
3 years ago
Colonist who belived in complete​
velikii [3]

Answer:

Natural right

Explanation:

That  is all what i know

5 0
3 years ago
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