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victus00 [196]
2 years ago
7

A metal block of mass 235 g rests at a point 2.8 m from the center of a horizontal rotating wooden platform. The coefficient of

static friction between the block and the platform is 0.437. The platform initially rotates very slowly but the rotation rate is gradually increasing. The acceleration of gravity is 9.8 m/s 2 . At what minimum angular velocity of the platform would the block slide away? Answer in units of rad/s.
Physics
1 answer:
Nuetrik [128]2 years ago
5 0

Answer:

ω=1.23 rad/sec

Explanation:

The block will start sliding when the centrifugal force exceeds the force of friction

We have Force of friction and centrifugal force are as under

F_{f}=\mu mg...........(i)\\\\F_{c.f}=m\omega ^{2}r................(ii)

Thus equating the 2 forces to get the minimum angular velocity we have

\fr\mu mg= m\omega^{2}r\\\\\therefore \omega =(\frac{\mu g}{r})^{1/2}

Applying values we get

ω=1.23 rad/sec

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Olenka [21]

Answer:

so angular velocity is 7.13128 sec−1

Explanation:

velocity v = 2.2 m/s

displacement s = 220 mm = 0.220 m

distance d = 510 mm = 0.510 m

to find out

angular velocity

solution

we know that

angular velocity will be velocity ( v)  / (displacement²  +  distance²)   .....1

now put all these value in equation 1 and we get angular velocity i.e.

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angular velocity = 2.2  / (0.22²  +  0.51²)

angular velocity = 2.2 / 0.3085

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6 0
3 years ago
A box is sliding down a ramp how many force vectors does the box have
vlada-n [284]
Well you’d have a force due to gravity, the normal force which will be perpendicular to the sources (meaning you’ll have components to this vector), and you’d have the force of friction opposing the motion of the box. I’m also assuming there’s no air resistance. In this case you’d have three vector forces.
6 0
3 years ago
A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
jeka57 [31]

Answer:

a. 3.1 m/s^2

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

mg sin \theta - \mu_k R = ma

where :

m = 6.0 kg is the mass  of the box

g = 9.8 m/s^2 the acceleration of gravity

\theta=39^{\circ}  is the angle of the slope

\mu_k = 0.40 is the coefficient of friction

R is the normal reaction  

a is the acceleration

- Along the perpendicular direction:

R-mg cos \theta =0

From the 2nd equation, we get an expression for the reaction force:

R=mg cos \theta

And substituting into the 1st equation, we can find the acceleration:

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2

6 0
3 years ago
A sound wave of frequency
Artyom0805 [142]

Answer:

0.283m

Explanation:

Speed (v) = wavelength × Frequency (f)

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Wavelength = 343/ 1210 = 0.283m

8 0
3 years ago
A 2.50 kg ball is attached to a 3.00 m bar and swung in a vertical circle. If the ball does not leave the circular loop, what mi
dezoksy [38]

Answer:

5.42 m/s

Explanation:

At minimum speed, the tension in the bar will be 0 when the ball is at the top of the arc, so the only force is gravity pulling down.

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∑F = ma

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v = √(gr)

v = √(9.8 m/s² × 3.00 m)

v = 5.42 m/s

6 0
3 years ago
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