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NARA [144]
3 years ago
7

In a control system, an accelerometer consists of a 4.63-g object sliding on a calibrated horizontal rail. A low-mass spring att

aches the object to a flange at one end of the rail. Grease on the rail makes static friction negligible, but rapidly damps out vibrations of the sliding object. When subject to a steady acceleration of 0.832g, the object should be at a location 0.450 cm away from its equilibrium position.
Required:
Find the force constant of the spring required for the calibration to be correct.
Physics
1 answer:
Masteriza [31]3 years ago
7 0

Answer:

8.4 N/m

Explanation:

m = Mass of block = 4.63 gm

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

x = Displacement of spring = 0.45 cm

a = Acceleration of subject = 0.832g

k = Spring constant

Force is given by

F=ma

From Hooke's law

F=kx

So

ma=kx\\\Rightarrow k=\dfrac{ma}{x}\\\Rightarrow k=\dfrac{4.63\times 10^{-3}\times 0.832\times 9.81}{0.45\times 10^{-2}}\\\Rightarrow k=8.4\ \text{N/m}

The force constant of the spring is 8.4 N/m.

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The sun, like all stars, releases energy through nuclear fusion. In this problem, you will find the total number of fusion react
sattari [20]

Answer:

Explanation:

DetaM=4 x 1.02875 - 4.002603

DetaM= 0.028697u

Using E= mc²

= 0.028697 x 1.49x*10^-10

= 4.2x10^-12J

4 0
3 years ago
A jet plane lands with a speed of 100 m/s and can
kiruha [24]

Answer:

a) t = 20 [s]

b) Can't land

Explanation:

To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.

a)

v_{f}=v_{i}-(a*t)

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = desacceleration = 5 [m/s^2]

t = time [s]

Note: the negative sign of the equation means that the aircraft slows down as it stops.

0 = 100 - 5*t

5*t = 100

t = 20 [s]

b)

Now we can find the distance using the following kinematics equation.

x -x_{o}=(v_{o}*t)+\frac{1}{2}*a*t^{2}

x - xo = distance [m]

x -xo = (0*20) + (0.5*5*20^2)

x - xo =  1000 [m]

1000 [m] = 1 [km]

And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land

4 0
3 years ago
A 8.4-mH inductor carries a current I = Imaxsin ωt, with Imax = 4.00 A and f = ω/2π = 60.0 Hz. What is the self-induced emf as a
Aleks04 [339]

Answer:

E= -3.166 cosωt   V

Explanation:

Given that

I = Imax sinωt

L= 8.4 m H

Imax= 4 A

f = ω/2π = 60.0 Hz

ω = 120π  rad/s

We know that self induce E given as

E=-L\dfrac{dI}{dt}

\dfrac{dI}{dt}= Imax \ \omega\ cos\omega t

E=-L\times Imax \ \omega\ cos\omega t

E=-8.4\times 120\times \pi \ cos\omega t

E= -3166.72 cosωt  m V

E= -3.166 cosωt   V

This is the induce emf.

3 0
3 years ago
A light ray transfers from more to less dense medium at certain condition , if we repeat this experiment by increasing the angle
evablogger [386]

Answer:

The correct option is;

Still constant

Explanation:

The relative refractive index ₁n₂ between the two medium can be as follows;

_1n_2 = \dfrac{The \ speed \ of \ light \ in \ medium \ 1}{The \ speed \ of \ light \ in \ medium \ 2}  = \dfrac{c_1}{c_2}

Therefore, given that the speed of light in medium 1 is constant and the speed of light on medium 2 is also constant, the relative refractive index ₁n₂ = c₁/c₂ is always constant.

4 0
3 years ago
How are the direction of the net force on an object and the direction of the object’s acceleration related?
Oksanka [162]

They're the same direction.

8 0
3 years ago
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