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Naddika [18.5K]
3 years ago
12

The coefficient of friction between a car and dry pavement is approximately 07, but the coefficient on wet pavement is 04 or

Physics
2 answers:
Brut [27]3 years ago
7 0
C
Because that’s the right answer !!!
diamong [38]3 years ago
4 0

Answer:

C. A car requires less applied force to move on wet pavement

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A projectile is launched with a velocity of 13.2m/s at an angle of 37 degrees above the horizontal. What is the horizontal compo
MrRissso [65]

Answer:

horizontal component = 10.54m/s

Explanation:

horizontal component = 13.2cos37°

horizontal component = 10.54m/s

8 0
3 years ago
• what is the typical distance between two adjacent pins on a 14-pin dual-in-line ic package?
muminat

A 14 pin dual-in-line IC package[14 DIL] is an integrated socket which is most popular form of IC package and has a wide range of application in digital electronics.

The 14-pin DIL has two pairs per side and each pair contains seven connecting pins.

The pairs of pins are arranged linearly one after another.The typical dimensions of width is 6.5 mm and the typical dimension of length is 18 mm.

we are asked to calculate the typical distance between two adjacent pins.

The typical distance between two adjacent pins is calculated as-

                                                                 Typical\ distance =\frac{dimensional\ length}{number\ of\ pins\ in\ each\ row}

                                    =\frac{18 mm}{7}

                                    = 2.5714 mm    [ans]                  

7 0
3 years ago
As a compass finds the northern direction which of the following is it aligning with
Anna [14]
It would be aligning with the north pole. 
8 0
3 years ago
Read 2 more answers
You travel 20.0 mph for 0.500 hour, 40.0 mph for 1.00 hour, 30.0 mph for 0.500 hour, and 50.0 mph for 2.00 hours. What is the av
Karo-lina-s [1.5K]
The answer is 41.25 mph.

The average speed (v) can be expressed as:
v = D/T  
where
D is total distance: D = d1 + d2 + d3 + d4
T is total time: T = t1 + t2 + t3 + t4

Now let's calculate distances using the formula: d = v * t:
d1 = v1 * t1 = 20.0 * 0.500 = 10 mi
d2 = v2 * t2 = 40.0 * 1.00 = 40 mi
d3 = v3 * t3 = 30.0 * 0.500 = 15 mi
d4 = v4 * t4 = 50.0 * 2.00 = 100 mi

Thus, the average speed is:
v= \frac{D}{T}= \frac{d1 + d2 + d3 +d4}{t1+t2+t3+t4} = \frac{10+40+15+100}{0.500+1.00+0.500+2.00}= \frac{165}{4} = 41.25 mph
4 0
3 years ago
An Airplane taxis onto the runway going at 10 m/s. If it can accelerate steady at 3
kirza4 [7]

Answer: 1333 m

the length of runway it will need is S = \frac{90^{2}-10^{2}  }{2.3}=1333 (m)

Explanation:

4 0
3 years ago
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