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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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Which structures main function is to produce food (sugar) in a plant
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Technically speaking, there are many "levels" of a plant for which this may hold true. I think the one you are referring to is the chloroplast. It takes in the light energy from the sun, water and carbon dioxide, and performs photosynthesis on them to produce sugar and oxygen. A leaf would also fit the description as this is a very general question.
8 0
3 years ago
Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3
VARVARA [1.3K]

Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

\lambda = 630 nm

now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

f = \frac{3 \times 10^8}{630 \times 10^{-9}}

f = 4.76 \times 10^{14} Hz

Part b)

Position of Nth maximum intensity on the screen is given as

y_n = \frac{n\lambda L}{d}

so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

n = 3

L = 1.4 m

0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

d = 3.48 \times 10^{-4} m

Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})

\theta = 0.311 degree

8 0
3 years ago
Describe the motion of an automobile on an east-west
Tanya [424]

Answer:

1. The automobile is traveling due east and is speeding up.

2. The car is traveling due east and is is slowing down.

3. The automobile is traveling due east at a constant speed.

4. The car is traveling due west and is slowing down.

5. The automobile is traveling due west and is speeding up.

6. The automobile is traveling due west at a constant speed.

7. The automobile is accelerating due east from rest.

8. The automobile is accelerating due west from rest.

Explanation:

The key to understanding this is:

When the acceleration and initial velocity of the automobile have the same sign (positive or negative) then the automobile is speeding up. Explained further, if acceleration and the initial velocity are both positive or they are both negative the automobile is speeding up but whenever they have opposite signs (that is acceleration is positive and initial velocity is negative or vice versa) the automobile is slowing down. When the acceleration is zero the automobile is maintaining a unform motion at a constant speed (the speed is not changing with time). The + or - sign indicates the direction of travel. In this case east is + and west is -. It is my pleasure answering this question. I hope you find it helpful. Thank you.

4 0
3 years ago
A conducting sphere with neutral charge is connected to the ground. A positively charged rod is moved near the sphere. While the
statuscvo [17]

Answer:

A negative

when the charged rod is brought closer to the sphere negative charges get induced on the surface of the sphere by gathering some electrons from the ground and the negative charges reamain on the sphere even after disconnecting it. here the ground acts as a reservoir of electrons.

5 0
3 years ago
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