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Vanyuwa [196]
2 years ago
14

A system consists of a disk rotating on a frictionless axle

Engineering
1 answer:
kakasveta [241]2 years ago
8 0

The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.

<h3>What is the angular momentum?</h3>

The angular momentum of the device earlier than and after the clay sticks can be the same.

Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.

  1. Li = Lf where;
  2. li is the preliminary second of inertia
  3. If is the very last second of inertia
  4. wi is the preliminary angular velocity
  5. wf is the very last angular velocity
  6. Li is the preliminary angular momentum
  7. Lf is the very last angular momentum

Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.

Read more about the frictionless :

brainly.com/question/13539944

#SPJ4

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In Josiah Johnson Hawes and Albert Sands Southworth, Early Operation under Ether, Massachusetts General Hospital the elevated vi
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C and A I think cause I don’t really remember this I done before it his to be C and A
4 0
3 years ago
The concrete canoe team does some analysis on their design and calculates that they need a compressive strength of 860 psi. They
vlada-n [284]

Answer:

874 psi

Explanation:

Given a sample mean (x') = 900,

and a standard error (SE) = 10

At 99% confidence, Z(critical) = 2.58

That gives 99% confidence interval as,

x' ± Z(critical) x SE = 900 ± 2.58 x 10

The value of the lower limit is,

900 - 25.8 = 874.2

≈ 874 psi

8 0
2 years ago
The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc
madreJ [45]

Answer:

Option B

116 ft/s^{2}

Explanation:

\theta=2 rev=2(2\pi)=4\pi

\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})

\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})

\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})

(2) (8\pi)= (\omega_f^{2}-4^{2})

\omega_f=8.14 rads/s

v=r\omega=1.75*8.14=14.245 ft/s

Centripetal acceleration =\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}

Tangential component=dr=2*1.75=3.5

Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}

5 0
3 years ago
Which statement about lean manufacturing is true when you compare it to mass production?
Len [333]
Where are the statements then bbs lol
6 0
2 years ago
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
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