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leonid [27]
3 years ago
6

The equation y = 2x + 3 represents the cost y (in dollars) of mailing a package that weighs x pounds.

Physics
1 answer:
White raven [17]3 years ago
6 0

Answer:

what is your question though

Explanation:

??? ??? ??? ??? ???

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Which one of the following is not an example of the wedge? <br> ax <br> chisel <br> nut <br> nail
Shkiper50 [21]
NUT...............................

6 0
3 years ago
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Una bola de acero rueda y cae por el borde de una mesa desde 4ft por encima del piso. Si golpea el suelo a 5ft de la base de la
ycow [4]

Answer:

     \large\boxed{\large\boxed{10ft/s}}

Explanation:

The question, translated, is:

  • <em>A steel ball rolls and falls off the edge of a table from 4ft above the floor. If you hit the ground 5ft from the base of the table, what was your initial horizontal velocity?</em>

<em />

<h2>Solution</h2>

<em />

This is a projectile motion, for which, the equations that you will need are:

    V_x_0= V_x=x\cdot t

     y=y_0+Vy_o\cdot t-g\cdot t^2/2

<u />

<u>1. Calculate the time that it takes the ball to fall 4ft</u>

    0=4ft-g\cdot t^2/2\\\\t^2=2\times 4ft/( 32.174ft/s^2)=0.24865s^2\\\\t=0.4986s

<u />

<u>2. Calculate the horizontal velocity:</u>

     V_x_0= V_x=x\cdot t\\\\V_x_0=5ft/0.4986s=10.027ft/s\approx 10ft/s

3 0
4 years ago
Suppose you average 75 MPH over the first halfof a drive, and your average speed for the entiredrive is 60 MPH. What was your av
alexandr402 [8]

Answer:

x = 45 MPH

Explanation:

given,

Average speed of the first half = 75 MPH

Average speed of entire ride = 60 MPH

Average speed of the second half = ?

let the average speed of the second half = x MPH

now,

average of entire ride is given as 60 mph so,

 \dfrac{75+x}{2} = 60

 75+x= 2\times 60

       75 + x = 120

         x = 120 -75

        x = 45 MPH

hence, the average speed of the second half comes out to be 45 MPH.

5 0
3 years ago
A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim
Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\   where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]

3 0
3 years ago
Which object will take the most force<br> to accelerate? *<br> 4 kg<br> 6 kg<br> 8 kg<br> 02 kg
r-ruslan [8.4K]

Answer:

I think it might be 8kg grams because it is bigger

5 0
3 years ago
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