Ask question

Ask question

All categories

2 years ago

6

The distance traveled by an object can be modeled by the equation d = ut + 0.5at2 where d = distance, u = initial velocity, t =

time, and a = acceleration. Solve this formula for a. Show all steps in your work. plz someone >~

1 answer:

Ostrovityanka [42]2 years ago

6 0

Given d = ut + 0.5at2, and assuming d or distance, u or initial velocity, and t or time are known, acceleration or a could be solved.

d – ut =0.5at2

(d – ut)/(t^2) = 0.5a

( d – ut)/[(0.5)(t^2)] = a

Hope this helps

You might be interested in

Determine whether each set of side lengths could be the sides of a right triangle.

kvv77 [185]

Answer:

i think both of them cant be

8 0

1 year ago

I need help pleasee!!!!

andre [41]

Answer:

Your answer is c

Step-by-step explanation:

the two inequalities are x =< 1 and x =>2 which is shown by the number line in option c

8 0

2 years ago

8. How much energy does a 12 V battery in your car use to move 2.5 C through the electrical

I am Lyosha [343]

The energy required is 30 watts second.

In order to accomplish work and to produce heat and light, energy must be transferred to a body or to a physical system. Energy is the quantitative attribute that does this. Energy is a preserved resource; according to the rule of conservation of energy, energy can only be transformed from one form to another and cannot be created or eliminated.

The voltage of a battery is 12 V and;

The charge is 2.5C.

Now, energy can be defined as the product of voltage and charge.

Therefore,

E = V × C

E = 12 × 2.5

E = 30 watt second

Hence, the energy needed by the 12 V battery to move 2.5 C through the electrical circuit is 30 watts second.

Learn more about energy here:

brainly.com/question/13881533

#SPJ9

6 0

1 year ago

Over the weekend, I bought a dress for $23.96, a pair of

Inga [223]

Answer:

4.94

Step-by-step explanation:

6 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

2 years ago