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viva [34]
3 years ago
13

A car starts from rest at a stop sign. It accelerates at 4.6 m/s^2 for 6.2 s , coasts for 2.1s , and then slows down at a rate o

f 3.3 m/s^2 for the next stop sign. How far apart the are the stop signs?
Physics
1 answer:
ki77a [65]3 years ago
5 0

Answer:

D = 271.54 m

Explanation:

given,

1. car accelerates at 4.6 m/s² for 6.2 s

2. constant speed for 2.1 s

3. slows down at 3.3 m/s²

distance travel for case 1

using equation of motion

 d_1 = u t +\dfrac{1}{2}at^2

 d_1 =\dfrac{1}{2}\times 4.6\times 6.2^2

      d₁ = 88.41 m

case 2

constant speed for 2.1 s now, we have to find velocity

v = u  + at

v = 0 + 4.6 x 6.2

v = 28.52 m/s

distance travel in case 2

d₂ = v x t

d₂ = 28.52 x 2.1 = 59.89 m

for case 3

distance travel by the car

v² = u² + 2 a s

final velocity if the car is zero

0² = 28.52² + 2 x (-3.3) x d₃

6.6 d₃ = 813.39

 d₃ = 123.24 m

total distance travel by the car

D = d₁ + d₂ + d₃

D = 88.41 + 59.89 + 123.24

D = 271.54 m

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kramer
To answer this question, we should know the formula for the terminal velocity. The formula is written below:

v = √(2mg/ρAC)
where
m is the mass
g is 9.81 m/s²
ρ is density
A is area
C is the drag coefficient

Let's determine the mass, m, to be density*volume.
Volume = s³ = (1 cm*1 m/100 cm)³ = 10⁻⁶ m³
m = (1.6×10³ kg/m³)(10⁻⁶ m³) = 1.6×10⁻³ kg
A = (1 cm * 1 m/100 cm)² = 10⁻⁴ m²

v = √(2*1.6×10⁻³ kg*9.81 m/s²/1.6×10³ kg/m³*10⁻⁴ m²*0.8)
<em>v = 0.495 m/s</em>
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What do you observe as you immerse the empty jug in a bucket which contains water? Explain why you observe this phenomenon. ​
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The jug drowns because the density of the jug is more than that of the density of water.

7 0
2 years ago
Read 2 more answers
Can someone help me?
Cloud [144]

What do we know that might help here ?

-- Temperature of a gas is actually the average kinetic energy of its molecules.

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Knowing just these little factoids, we realize that as a gas gets hotter, the average speed of its molecules increases.

That's exactly what Graph #1 shows.

How about the other graphs ?

-- Graph #3 says that as the temperature goes up, the molecules' speed DEcreases.  That can't be right.

-- Graph #4 says that as the temperature goes up, the molecules' speed doesn't change at all.  That can't be right.

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An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

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f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

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f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0
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