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3 years ago

What is 1wfqWR x5nrznzrnzrn

Mathematics

2 answers:

Nadusha1986 [10]3 years ago

8 0

Answer:

it equaklsb=5

Step-by-step explanation:

youbsggcc t dxtdfd6r

Alborosie3 years ago

7 0

Answer:

could you type that again? I don't understand

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Sue has 18 sweets

Nikolay [14]

Answer:Find out the expressions for the number of sweets sue and tony have now .

To proof

As given

Sue has 18 sweets. tony has 18 sweets.sue gives tony x sweets

thus

sue sweets = 18 - x

Tony sweets = 18 + x

As given

Sue then eats 5 sweets. Tony then eats half of his sweets

sue sweets becomes = 18 -5 -x

= 13 - x

Hence proved

Step-by-step explanation:

6 0

3 years ago

PLEASE HELP ASAP!!!!! i’ll mark brainlest

hoa [83]

Answer:

It would be hundreds

Step-by-step explanation:

I hope it is right :I

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

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

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