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Rasek [7]
3 years ago
14

Does the asteroid belt revolve around the sun?

Physics
1 answer:
Vanyuwa [196]3 years ago
3 0

These small rocky objects orbiting the sun have no atmosphere and are traces from the formation of the solar system. Thousands of asteroids can be found revolving around the sun in an “asteroid belt", an area located between the orbits of Mars and Jupiter.


brainliest please?

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The world’s largest wind turbine has blades that are 80 m long and makes 1 revolution every 5.7 seconds. What is the velocity fo
klemol [59]

Answer:

The free end of the blade has a tangential velocity of about 88.19 m/s

Explanation:

The angular velocity of the blades is  2 \pi /5.7\,\,rad/sec

since the blades are 80 m long, then the tangential velocity of the free end of the blade is:

v_{tan} \approx 88.19\,\,m/s

3 0
3 years ago
An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef
navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0
3 years ago
A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s
Jlenok [28]

Answer:

The applied torque is 3.84 N-m.      

Explanation:

Given that,

Moment of inertia of the wheel is 2\ kg-m^2

Initial speed of the wheel is 0 (at rest)

Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m

So, the applied torque is 3.84 N-m.

4 0
3 years ago
In DC motor, the split rings are made of <br> A) steel B) copper C) wood D) glass
USPshnik [31]

I think is

(B) Copper

8 0
3 years ago
A pencil sharpener has a handle with a radius of 1.5 cm. The sharpening mechanism has a radius of 0.5 cm. What is the IMA of the
abruzzese [7]
1.5 / 0.5 = 3 I believe this is the right answer
8 0
3 years ago
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