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3 years ago

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Abigail is going for a hike after six KM she gained 800 M in elevation she stopped for a break then turns around and returned th

e way she came how many kilometers did Abigail hike in total and what was her total change in elevation when she finished the height

2 answers:

Ivan3 years ago

6 0

12 km and the difference is 800

Vika [28.1K]3 years ago

3 0

Answer:

12 Km and the difference was 800 m

Step-by-step explanation:

If she stopped 6 Km in and turned around it would give her another 6 Km to get back. When she climbed she gained 800 M in elevation because this is the amount she reached from her starting point.

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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

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

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

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

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

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

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

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

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

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