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FrozenT [24]
3 years ago
9

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has

a length of 51 cm and is wrapped around the top at a place where its radius is 1.8 cm. The thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of the string, thereby unwinding it and giving the top an angular acceleration of 10 rad/s2. What is the final angular velocity of the top when the string is completely unwound
Physics
1 answer:
AlladinOne [14]3 years ago
3 0

Given Information:

Angular displacement = θ = 51 cm = 0.51  m

Radius = 1.8 cm = 0.018 m

Initial angular velocity = ω₁ = 0 m/s

Angular acceleration = α = 10 rad/s ²

Required Information:

Final angular velocity = ω₂ = ?

Answer:

Final angular velocity = ω₂ = 21.6 rad/s

Explanation:

We know from the equations of kinematics,

ω₂² = ω₁² + 2αθ

Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.

ω₂² = (0)² + 2αθ

ω₂² = 2αθ

ω₂ = √(2αθ)

We know that the relation between angular displacement and arc length is given by

s = rθ

θ = s/r

θ = 0.51/0.018

θ = 23.33 radians

finally, final angular velocity is

ω₂ = √(2αθ)

ω₂ = √(2*10*23.33)

ω₂ = 21.6 rad/s

Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.

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horrorfan [7]

Answer:

2271.16N/C  upward

Explanation:

The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.

First we determine the downward weight using

W=mg\\g=9.81m/s^{2}

Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be

W=0.00382kg*9.81m/s^{2}

W=0.0375N

To calculate the electric field,

E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C

Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is  repelling it, Hebce we can conclude that the electric field lines are upward.

Hence the magnitude of the electric force is 2271.16N/C and the direction is upward

4 0
3 years ago
How do sea surface temperatures affect evaporation rate?
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7 0
3 years ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

8 0
3 years ago
I dont know how to do this at all please help
worty [1.4K]
Wow !  I understand your shock.  I shook and vibrated a little
when I looked at this one too.

The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.

"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?

Acceleration of gravity is

                           G  ·  M / R²

      =  (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

      =  (6.67 x 10⁻¹¹  ·  1.1 x 10³¹ / 4 x 10⁶)      (N) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴           (kg · m / s²) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴            m / s²      

That's about  1.87 x 10¹³  times the acceleration of gravity on
Earth's surface.

In other words, if I  were standing on the surface of that neutron star,
I would weigh  1.82 x 10¹² tons, give or take.     
3 0
3 years ago
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