Ask question

Ask question

All categories

2 years 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.

1 answer:

Semenov [28]2 years 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'

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

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

You might be interested in

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

5 0

2 years ago

Often what one expects to see influences what is perceived in the surrounding environment. Please select the best answer from th

Stella [2.4K]

Hello <span>Andijwiltbank
</span>

Question: <span>Often what one expects to see influences what is perceived in the surrounding environment. True or False?

Answer: True

Reason: What we observe about the environment decides what we believe about it and how we react.

Hope This Helps :-)
-Chris</span>

8 0

2 years ago

Read 2 more answers

The amount of light that undergoes reflection or transmission is demonstrated by how bright the reflected or transmitted ray is.

melamori03 [73]

If you subscribe I’ll answer QF Aotrx

8 0

2 years ago

If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 14m/s, how far from the base of the dorm should

liubo4ka [24]

Emily throws the ball at 30 degree below the horizontal

so here the speed is 14 m/s and hence we will find its horizontal and vertical components

$v_x = 14 cos30 = 12.12 m/s$

$v_y = 14 sin30 = 7 m/s$

vertical distance between them

$\delta y = 4 m$

now we will use kinematics in order to find the time taken by the ball to reach at Allison

$\delat y = v_y *t + \frac{1}{2} at^2$

here acceleration is due to gravity

$a = 9.8 m/s^2$

now we will have

$4 = 7 * t + \frac{1}{2}*9.8 * t^2$

now solving above quadratic equation we have

$t = 0.44 s$

now in order to find the horizontal distance where ball will fall is given as

$d = v_x * t$

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

$d = 12.12 * 0.44 = 5.33 m$

so the distance at which Allison is standing to catch the ball will be 5.33 m

8 0

2 years ago

8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres

Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

$\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}$

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

$\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)$

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

$\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T$

hence, B_o = 1.013μT

6 0

2 years ago