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Goryan [66]
1 year ago
15

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

hin rim with radius 0.54 cm , connected to the balance staff by thin spokes of negligible mass. The total mass of the balance wheel is 0.95 g .Part AWhat is the moment of inertia of the balance wheel about its shaft?Express your answer using two significant figures.Part BWhat is the torsion constant of the hairspring?Express your answer using two significant figures.
Physics
1 answer:
Semenov [28]1 year ago
3 0

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

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Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})

h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}

2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0
2 years ago
A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the
egoroff_w [7]

The acceleration of the jet is 28.1 m/s^2

Explanation:

Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0

v = 75 m/s

s = 100 m

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{75^2-0}{2(100)}=28.1 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
2 years ago
The diagram shows Ned’s movement as he left his house and traveled to different places throughout the day. Which reference point
aleksley [76]

The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

6 0
2 years ago
Read 2 more answers
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
2 years ago
Plug variables expressed in SI units in the kinematic equation given in article: a = -v0^2/(2sg). What value of g you get as exp
Elanso [62]

Answer:

1) acceleration is increased by a factor of four 4X

2) the acceleration increases a factor of 2X

3) the correct answer of 400g

Explanation:

This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.

           v² = v₀² + 2 a x

           0 = v₀² + 2 a x

           a = - v₀² / 2 x

           

In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities

         a / g = - v₀² / (2 x g)

Let's answer the different questions about this equation

1. The initial velocity is doubled, how much the acceleration is worth

           

       a/g = - (2v₀) 2 / 2xg

       a = 4 (-v₀² / 2xg) g

acceleration is increased by a factor of four 4X

2. if the stopping distance is reduced by 2, that is, x = x₀ / 2

we substitute

        a/g = (- v₀² / 2g) 2/x

         

        a =2  (-v₀² / 2x₀g)  g

       

therefore the acceleration increases a factor of 2X

3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is

x = 5cm = 0.05m

we calculate the acceleration

        a / g = - 20² / (2 0.05)

        a / g = - 4000 / g

        a / g = - 4000 / 9.8 = 408

        a = 408 g

the correct answer of 400g, the value matches exactly if g = 10 m / s2 is taken

6 0
2 years ago
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