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3 years ago

How to solve a torque problem

Physics

1 answer:

podryga [215]3 years ago

6 0

<span>you must first select an axis of rotation about which to calculate moment arms and torques. </span>

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Which sentence describe newton's third law?

strojnjashka [21]

C. Forces are always in pairs

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3 years ago

A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 5.12 g coins stacked over the 24.2 cm mark, th

pashok25 [27]

Answer:

The mass of the meter stick is $1.66054054054g$

Explanation:

Here since the meter stick is in equilibrium position its net torque and net force should be equal to zero

Since at the begging the meter stick is balanced at center of the meter stick that means its center of mass should be present at $50.0cm$

Now lets consider the later case where stick is balanced by two 5.12 g coins .

Here torque due to two coins = $t_{c} = 5.12\times(27.8-24.2)\times2\times9.8$

Torque due to weight of meter stick = $t_{m}$ = $m\times(50-27.8)\times9.8$

where m = mass of the meter stick

Here $t_{c} = t_{m}$ .

Upon equating we will be getting mass of the meter stick = $1.66054054054g$

4 0

3 years ago

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

Anna35 [415]

This question is incomplete

Complete Question

Three equal point charges are held in place as shown in the figure below

If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.

A) F1=2F2

B) F1=3F2

C) F1=4F2

D) F1=9F2

Answer:

D) F1=9F2

Explanation:

We are told in the question that there are three equal point charges.

q, Q1, Q2 ,

q = Q1 = Q2

From the diagram we see the distance between the points d

q to Q1 = d

Q1 to Q2 = nd

Assuming n = 2

= 2 × d = 2d

Sum of the two distances = d + 2d = 3d

F1 is the force on q due to Q1 and

F2 is the force on q due to Q2,

Since we have 3 equal point charges and a total sum of distance which is 3d

Hence,

F1 = 9F2

6 0

4 years ago

Light from a laser strikes a diffraction grating that has 5500 lines per centimeter. The central and first-order maxima are sepa

WINSTONCH [101]

.Answer:

491.4 nm

Explanation:

The distance between central and first maxima is,

$y=0.455m$

And the distance between screen abnd grating is,

$L=1.62 m$

Now the angle can be find as,

$tan\theta=\frac{y}{L} \\\theta=tan^{-1}(\frac{0.455}{1.62}) \\\theta=15.68^{\circ}$

Now the grating distance is,

$d=\frac{1}{5500} cm\\d=1.82\times 10^{-6}m$

Now with m=1 condition will become,

$\lambda=dsin\theta$

So,

$\lambda=1.82\times 10^{-6}m\times sin(15.68^{\circ})\\\lambda=1.82\times 10^{-6}m\times 0.270\\\lambda=491.4\times 10^{-9}m\\\lambda=491.4 nm$

Therefore the wavelength of laser light is 491.4 nm.

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3 years ago

A positive symptom of schizophrenia would be

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Answer:

i will like to know the answer but idk

Explanation:

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3 years ago

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