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Verizon [17]
3 years ago
6

Why VR is always less than 1 in Third class lever?​

Physics
1 answer:
Sav [38]3 years ago
3 0
Mechanical advantage of a class 3 lever is always less than 1 because the load arm is always longer than effort arm. This helps by reducing the applied force(effort).that’s your answer hope this helps you!
Have a great day!
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SOMEBODY PLEASE HELP!!! Indicate the reasons why the centripetal acceleration (and centripetal force) always point to the center
maxonik [38]
Well we know the correct answer cannot be "a" bcause velocity is tangent to the circlular path of an object experienting centripical motion. Velocity DOES NOT point inward in centripical motion.

we know the correct answer cannot be "b" because "t" stands for "time" which cannot point in any direction. so, time cannot point toward the center of a circle and therefore this answer must be incorrect.

I would choose answer choice "c" because both force and centripical acceleration point toward the center of the circle.

I do not think answer choice "d" can be correct because the velocity of the mass moves tangent to the circle. velocity = (change in position) / time. Therefore, by definition the mass is moving in the direction of the velocity which does not point to the center of the circle.

does this make sense? any questions?

3 0
3 years ago
The 94-lb force P is applied to the 220-lb crate, which is stationary before the force is applied. Determine the magnitude and d
Makovka662 [10]

Answer

given,

force = 94 lb

weight of crate = 220 lb

Assuming the static friction be equal = 0.47

                       kinetic friction = 0.36

Maximum force applied to move the object is when object is just start to move.

F = μ N

F = 0.47 x 220

F = 103.4 lb

As the frictional force is more than applied then the object will not move.

so, the friction force will be equal to the force applied on the object that is equal to 94 lb.

hence, the direction of force will left.

8 0
3 years ago
A 50.0 kg woman climbs a flight of stairs 6.00 m high in 15.0 s. How much power does she use.
WITCHER [35]

Her weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

Work = (weight) · (height) = (50kg) · (9.8 m/s²) · (6 m)

Power = (work) / (time) = (50kg) · (9.8 m/s²) · (6 m) / (15 s)

Power = (50 · 9.8 · 6 / 15) · (kg · m² / s³)

Power = 196 (kg · m / s²) · (m) / s

Power = 196 Newton-meter/second

<em>Power = 196 watts</em>

6 0
3 years ago
Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus
poizon [28]

Answer:

For destructive interference phase difference is

(2n+1)\pi where n∈ Whole numbers

Explanation:

For sinusoidal wave the interference affects the resultant intensity of the waves.

In the given example we have two waves interfering at a phase difference of \frac{\pi}{4} would lead to a constructive interference giving maximum amplitude at at the RMS value of the amplitude in resultant.

Also the effect is same as having a phase difference of  ( \frac{\pi}{4} + 2\pi) because after each 2π the waves repeat itself.

<em>In case of destructive interference the waves will be out of phase i.e. the amplitude vectors will be equally opposite in the direction at the same place on the same time as shown in figure.</em>

They have a phase difference of \pi or which is same as (2\pi+\pi)

Generalizing to:

a phase difference of (2n+1)\pi where n∈ {W}

{W}= set of whole numbers.

3 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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