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4 years ago

Please answer this correctly as soon as possible as it’s due today

Mathematics

2 answers:

REY [17]4 years ago

8 0

Answer:

0.079mm2

Step-by-step explanation:

1/4 x 3.14x 1mm x 1mm= 0.079mm2( correct to hundredth place)

= 0.0785mm2

Andrews [41]4 years ago

7 0

Answer:

.79 mm^2

Step-by-step explanation:

First find the area of the circle

A = pi r^2

A = 3.14 (1) ^2

A = 3.14

Now we have 1/4 of the circle

The area is 1/4 of the full area

1/4 *3.14

.785

Rounding to the nearest hundredth

.79 mm^2

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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

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

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