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Oduvanchick [21]
3 years ago
9

A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi

nish line?
Physics
1 answer:
Ahat [919]3 years ago
6 0
<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.8 s    

        Displacement, s = 1/4 mi =    400 meters

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.8 + 0.5 x a x 6.8²

                      a = 17.30 m/s²

Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.8 s

      Acceleration, a = 17.30 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.30 x 6.8

                      v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

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Answer:

i. -4m

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Explanation:

The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)

Total distance = 8m going east + 8m back to origin + 4m west = 20m

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3 years ago
The angle of elevation from a point on the ground to the top of a pyramid is 37 degrees 50​' The angle of elevation from a point
Serjik [45]

Answer:

The height of the pyramid is approximately 104 Ft. See the graphic attached.

Explanation:

First, you have to plot to realize that you have two rectangle triangles, formed by the different elevation points of view. From there you can have a system of two equations, with two unknown values.

Equation (1)

tan 37^{o}50'= \frac{Ph}{x} \\\\Ph=tan 37^{o}50'  x= 0.7766 x

Equation (2)

tan 18^{o}10'=\frac{Ph}{x+183} \\\\Ph=[tan 18^{o}10'][x+183]=[0.3281][x+183]

Matching (1) and (2)

0.7766 x=0.3281(x+183)\\\\(0.7766-0.3281)x=60.05\\\\x=\frac{60.05}{0.4485} =133.89ft

replacing x value in (1)

Ph=0.7766*133.89=103.9789\\Ph = 104 ft

5 0
3 years ago
If the velocity of blood flow in the aorta is normally about 0.32 m/s, what beat frequency would you expect if 4.40-MHz ultrasou
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Answer:

The beat frequency is 0.0019 MHz.

Explanation:

Given that,

Velocity = 0.32 m/s

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We need to calculate the frequency

Case (I),

Observer is moving away from the source

Using Doppler's effect

f'=\dfrac{v-v'}{v}f

Where, v' = speed of observer

Put the value into the formula

f'=\dfrac{1540-0.32}{1540}\times4.40

f'=4.399\ MHz

Case (II),

Cell is as the source of sound of frequency f' and it moving away from the observer.

Using formula of frequency

f''=\dfrac{v-v_{s}}{v+v_{s}}\times f

f''=\dfrac{1540-0.32}{1540+0.32}\times4.399

f''=4.3971\ MHz

We need to calculate the beat frequency

\Delta f= f'-f''

\Delta f=4.399-4.3971=0.0019\ MHz

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4 0
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Alexxx [7]
<span>5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
4 0
3 years ago
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