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olchik [2.2K]
3 years ago
5

15 points for an answer for this question

Mathematics
1 answer:
sasho [114]3 years ago
5 0

Answer:

Is it Khan Academy? You do 5.8 - 2.7 = 3.1

Step-by-step explanation:

Hope that helps!

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Solve for x in the equation x2 +11x+121/4=125/4
aksik [14]

Answer:

x= −11 2 + 5 2 √5 or x= −11 2 + −5 2 5

Step-by-step explanation:

4 0
3 years ago
In 10 s, 200 bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mass
Dmitry_Shevchenko [17]

\large\bf{\underline{Answer:}}

\large\bf{a) \triangle  p_{1s} =  - 120 \: kgm {s}^{ - 1} }

\large\bf{b) F = -120N}

\large\bf{c) Pressure=40.10\times 10^5 pa }

__________________________________________

\large\bf{\underline{In\: this\: problem\:we\:have:}}

  • \bf{N = 200\: bullets}
  • \bf{M= 5\times 10^{-3}kg}
  • \bf{V= 1200\:{ms}^{-1}}

❒ To find the change in momentum for bullets , we need to remember the momentum p of a bullet is equal to product of mass and speed

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼p_{1}= mv}

❒ This means , that change in momentum for one bullet will be equal to

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle P_{1} = mv_{f} - mv_{i}}

\large\bf{where\:v_{f}=0}

Total change in momentum for the bullet in 10 sec is equal to product of change in momentum for one bullet and number of bullets hit the wall in 10 sec

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle P_{10s} = N\triangle P_{i}}

<h3>❒<u> </u><u>Note </u><u>:</u><u>-</u></h3>

Change in momentum given is the change of momentum in 10 sec is 10 times less

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p_{1s} = \frac{N\triangle p_{i}}{10}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p_{1s}=\frac{200.(mv_{f}-mv_{i}}{10}}

\large\bf{as\:said,v_{f}=0}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p_{1s}=\frac{-200.mv_{i}}{10}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p_{1s}=\frac{200.5\times 10^{-3}kg.1200ms^{-1}}{10}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p_{1s}=-1200\:Kgms^{-1}}

__________________________________________

<h3>b) to find average force F on the wall we must remember that in general case force us the change of momentum in time :</h3>

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼F =\frac{\triangle P}{\triangle t}}

Total change of momentum of bullets in 10 sec

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p =N\triangle p_{i} }

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p= N(mv_{f}-mv_{i})}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p= -N mv_{i}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p= -200.5\times 10^{-3}.1200ms^{-1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼\triangle p= -1200kgms^{-1}}

❒ We can find total force exerted in the wall in 10sec by dividing the momentum of bullet with 10 sec

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼F = \frac{\triangle p}{\triangle t}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼F = \frac{-1200}{10}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼F = -120N}

__________________________________________

<h3>c) To find average pressure :</h3>

\large\bf{area = 3\times 10^{-4}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼P=\frac{|F|}{A} }

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼P=\frac{-120}{3\times 10^{-4}}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟼P=40\times 10^4 Pa}

7 0
3 years ago
What is the constant of proportionality shown on the graph?
zavuch27 [327]

Answer:

The constant of a proportionality in a graph of a proportional relationship is the constant ratio of y to x (the slope of the line

6 0
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What number is multiplied to -1 and then adds to -5
Kryger [21]
Y: is the number

[Y*(-1)]+(-5)
(-Y)+(-5)
-Y-5

<span>I hope it helps you :)</span>

4 0
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Set it up as an equation.
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