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nikitadnepr [17]
2 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]2 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.

You might be interested in
A standard baseball has a circumference of approximately 23 cm . if a baseball had the same mass per unit volume as a neutron or
Svetlanka [38]

<u>Answer:</u>

 Mass of base ball  m_b=3.992*10^{14}kg  

<u>Explanation:</u>

  Circumference of baseball = 2πr = 23 cm

  So radius of baseball = 3.66 cm = 3.66*10^{-2} m

   Mass per unit volume of baseball = Mass per unit volume of neutron or proton.

   Mass of proton = 10^{-27} kg  

   Diameter of proton = 10^{-15} m

   Radius of proton =  5*10^{-16} m

   Volume of ball = \frac{4}{3} \pi r^3

   Now substituting all values in Mass per unit volume of baseball = Mass per unit volume of neutron or proton.    

         \frac{m_b}{\frac{4}{3}\pi *(3.66*10^{-2})^3} =\frac{10^{-27}}{\frac{4}{3}\pi *(5*10^{-16})^3}

         \frac{m_b}{(3.66*10^{-2})^3} =\frac{10^{-27}}{(5*10^{-16})^3}  

         m_b=3.992*10^{14}kg

       So mass of base ball  m_b=3.992*10^{14}kg              

5 0
2 years ago
Read 2 more answers
The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe
Novosadov [1.4K]

Answer:

a) a= 8.33 m/s²,    T = 12.495 N , b)    a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

            fr₁ + fr₂ = ma

The equation for the force of friction is

          fr = μ N

Y Axis

         N - W₁ –W₂ = 0

         N = W₁ + W₂

         N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

        N = 2 m g

We replace

           μ₁ 2mg + μ₂ mg = m a

          a = (2μ₁ + μ₂) g

          a = (2 0.30 + 0.25) 9.8

          a= 8.33 m/s²

Let's look for cable tension with beam 2

          T = m₂ a

          T = 1500 8.33

          T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

           fr = m₂ a₂

           N- w₂ = 0

          N = W₂ = mg

          μ₂ mg = m a₂

          a = μ₂ g

          a = 0.25 9.8

          a = 2.45 m / s²

4 0
2 years ago
23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
2 years ago
Before the student releases the cart by cutting the tinsel string, what forces are acting on the cart?
Naya [18.7K]

Answer: TENSION and WEIGHT

Explanation:

Force experienced by the spring is called TENSION while the WEIGHT is the gravitational pull on the body towards the earth surface. Therefore the forces acting on the cart are TENSION and WEIGHT(weight acts downwards (along negative y-axis) while the TENSION upward(along positive y-axis).

5 0
2 years ago
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