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Svetlanka [38]
3 years ago
10

A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an

gles of (a) 30∘, (b) 30 rad, and (c) 30 rev, respectively? (a) Wheel rotates 30∘. 3.09 m ( ± 0.2 m) (b) Wheel rotates 30 rad. 18.226 m ( ± 20 m) (c) Wheel rotates 30 rev. 1112.1 m ( ± 20 m)
Physics
1 answer:
posledela3 years ago
8 0

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle \theta=30°

We need to calculate the angle in radian

\theta=30\times\dfrac{\pi}{180}

\theta=0.523\ rad

We need to calculate the path length

Using formula of path length

Path\ length =angle\times radius

Path\ length=0.523\times5.9

Path\ length =3.09\ m

(b). Angle = 30 rad

We need to calculate the path length

Path\ length=30\times5.9

Path\ length=177\ m

(c). Angle = 30 rev

We need to calculate the angle in rad

\theta=30\times2\pi

\theta=188.4\ rad

We need to calculate the path length

Path\ length=188.4\times5.9

Path\ length =1111.56\ m

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

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<u><em>on  flow properties and free-flowing and cohesive. </em></u>

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the power Free flowing powders do not cling together, as cohesive powders stick to each other and form that do not disperse well during mixing

6 0
3 years ago
Tire or false
spayn [35]
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Newton's law of gravity was inconsistent with Einstein's special relativity because
Lubov Fominskaja [6]

Answer:

Mass and thus force depends on the reference frame chosen

Explanation:

This can be explained as Newton's law of gravity provides action which are instantaneous at a distance and involves the evaluation of all the quantities at present time or at the instant they occur.

If the body undergoes a change in its mass distribution there will be an immediate change in its gravitational force without any lag.

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In order to observe Newton's law of gravity all the observer's in different reference frames must observe the same phenomena which could only be possible if time were absolute and in special relativity, time is not absolute.

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3 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2
lys-0071 [83]

Answer: sheet of charge

Explanation:

a )

Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.

b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge.   The equi- potential surface is always perpendicular to electric field.

C ) Electric field which is almost uniform near the sheet of charge is equal t the following

E = σ / ε₀ where  σ is charge density of surface and  ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²

E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²

= .3389 x 10³

= 338.9 V / m

spacing between 1 V

= 1 / 338.9 m

= 2.95 X 10⁻3 m

= 2.95 mm.

3 0
3 years ago
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