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DiKsa [7]
2 years ago
12

a tire is rolling along a road, without slipping with a velocity v. a piece of tape is attached to the tire. When the tape is op

posite the road (at the top of the tire), its velocity with respect to the road is
Physics
1 answer:
Kryger [21]2 years ago
4 0

Answer:

The right solution will be the "2v".

Explanation:

For something like an object underneath pure rolling the speed at any point is calculated by:

⇒  v_{rolling}=v_{translational}+v_{rotational}

Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:

⇒  \omega=\frac{v_{translational}}{r}

Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:

⇒  v_{rotational}=\omega \times r

On substituting the estimated values, we get

⇒                   =\frac{v_{translational}}{r} \times r

⇒                   =v_{translational}

So that the velocity will be:

⇒  v_{rolling}=v+v

⇒              =2v

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A 60.0-kg mass person wishes to push a 120-kg mass box across a level floor. The coefficient of static friction between the pers
Aneli [31]

Answer:

μ = 0.350

Explanation:

For the person to able to move the box, the force exerted by the person on the box must equal the force exerted by the box:

F_{p} = F_{b}

In this case, force can be calculated as a product of mass (m) by the acceleration of gravity (g) and the coefficient of static friction (μ):

m_{p}*g*\mu_{p}=m_{b}*g*\mu_{b}\\m_{p}*\mu_{p}=m_{b}*\mu_{b}\\60*0.7=120*\mu_{b}\\\mu_{b}= 0.35

Therefore, for the person to be able to push the box horizontally, the coefficient of static friction between the box and the floor should not be higher than 0.350.

8 0
2 years ago
Ancient Greek philosophers spent lots of time thinking about science and imaging explanations for the natural world. What part o
Illusion [34]

Answer:

Testing

Explanation:

Ancient Greek philosophers lived with the ideology to simply contemplate life. This means that their whole life revolved around thinking and questioning everything. This would include creative thinking, because they would sometimes come up with theories which require creativeness. They would often debate with their friends as to why their theory should be accepted or what their opinions were on the matter. More often than not, they argued a lot, and many philosophers went against some powerful people in the community and some were even sentenced to death.

The main process they didn't/couldn't do was the testing. They could never test certain theories because they did not have the means to.

4 0
2 years ago
You are learning about energy transformations in science class. Mel and Sam's built this set-up to see if light energy could be
Allisa [31]
I think it might be heat energy. light transforms into heat energy
5 0
2 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
2 years ago
a projectile is launched straight up at 141 m/s . How fast is it moving at the top of its trajectory? suppose it is launched upw
BARSIC [14]

The velocity of projectile has 2 components, horizontal component vcosθ and vertical component vsinθ, where v is the velocity of projection and θ is the angle between +ve X-axis and projectile motion.

In case 1, θ = 90⁰

    So horizontal component is vcos90 = 0

         Vertical component at maximum height = 0

So velocity at maximum height = 0 m/s


In case 2, θ = 45⁰

    So horizontal component is 141cos45 = 100m/s

         Vertical component at maximum height = 0

So velocity at maximum height = 100 m/s


7 0
2 years ago
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