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nikitadnepr [17]
3 years ago
11

Tech A says that tires are marked with a date code listing the date the tires should be discarded. Tech B says that any tires wi

th a three-digit date code should be discarded because they are too old to be safely in service. Who is correct?
Physics
1 answer:
9966 [12]3 years ago
4 0

Tech B is saying correct.

<u>Explanation:</u>

Yes, tires are provided with date codes but those date codes are correlated with manufacturing or production dates and not with the date of discarding. But it is difficult to understand or decode the date code. The tires with 3 digit code represent very old tires while the tires with 4 digit code are made after 1999.

In the four-digit date code, the first two digits indicate the week during which that tire was produced and the last two digits indicate the year of production. So the statement said by Tech A is wrong and the statement said by Tech B is correct.

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Find the acceleration of the blocks when the system is released. The coefficient of kinetic friction is 0.4, and the mass of eac
grin007 [14]

Answer:

a = 4.9(1 - sinθ - 0.4cosθ)

Explanation:

Really not possible without a complete setup.

I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g

                                     F = ma

mg - mgsinθ - μmgcosθ = (m + m)a

      mg(1 - sinθ - μcosθ) = 2ma

      ½g(1 - sinθ - μcosθ) = a

maximum acceleration is about 2.94 m/s² when θ = 0

acceleration will be zero when θ is greater than about 46.4°

8 0
3 years ago
Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through
liraira [26]

Answer:

Angular acceleration = 23.68 rad / s²

Explanation:

Given that,

acceleration = 9m/s²

Therefore acceleration of string is 9m/s²

since string is constant in length

cylinder of radius 38.0 cm = 0.38m

Angular acceleration = a / r

Angular acceleration = 9 / 0.38

                                   = 23.68 rad / s²

Angular acceleration = 23.68 rad / s²

4 0
3 years ago
4. Explain how states of matter change in regards to<br> a. Temperature-<br> b. Pressure-
Alina [70]
Temperature can change the state from solid to liquid causing it to melting, liquid to gas causing vaporization or a solid to a gas causing sublimation. Pressure alone cannot change the state of matter.
5 0
3 years ago
Help me with this???
Vikentia [17]

Yo sup??

Average velocity=total distance covered/total time taken

total distance covered=4 + 8=12 miles

total time taken=6 hours

Therefore

average velocity=12/6

=2 miles/hour

Hope this helps

8 0
3 years ago
A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.
Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = 0.5mv^{2}

           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
  • when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = 0.5x25x5.6^{2} - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

5 0
3 years ago
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