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adoni [48]
2 years ago
11

A group of friends wants to go to the amusement park. They have no more

Physics
2 answers:
Mumz [18]2 years ago
5 0

Answer:

p ≤ 8

Explanation:

Let's go over what we know that can help us build an inequality:

we know that these friends can spend no <em>more </em>than $325, meaning that they can either spend exactly $325, or they could spend less. This is our first clue, that one side of the inequality will be 325, and that the inequality sign will be <em>less than or equal to </em>( ≤ ) :

amount spent ≤ 325

Now, we want to figure out how many people can go (p)--but there's something we need to consider first--the parking.

Assuming that all friends take one car, we know we will have 11 dollars less to spend on admission (you can write this two ways).

amount spent on tickets + 11 ≤ 325

or, we can simplify it:

amount spent on tickets + 11 ≤ 325

                                          - 11     -11

amount spent on tickets ≤ 314

(I will be using the second way, you could always plug it back into the first way though)

now, we know that each "p" that we include will be multiplied by 39.25

         <em>Why?  Imagine we have one p (one person), they will cost 39.25, so we could write that as (p · 39.25) ,  ( 1· 39.25 ) -- we use × because that's how many </em><em>times</em><em> we're adding the number.  If we have two p, we could write it as (2p · 39.25)--there's an easier way to do this though</em>

<em />

So, we can write "p" to stand for all possible values of p possible,

p · 39.25 ≤ 314                  (we write this slightly differently):

39.25p ≤ 314          

Now, let's simplify!

   39.25p ≤ 314          

÷ 39.25     ÷ 39.25  (divide by 39.25 to isolate p)

          p   ≤   8

So, we know that the number of friends admitted must be less than or equal to 8, expressed as:

p ≤ 8

hope this helps! let me know if you need clarification on anything :)

kirill [66]2 years ago
5 0

Answer:

4x + 3

Explanation:

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koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

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4 0
3 years ago
A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

\theta=69.44^{\circ}

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

T=\dfrac{3\times9.8}{2\times\cos69.44}

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3 years ago
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In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

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