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Lana71 [14]
2 years ago
11

Three objects of the same mass begin their motion at the same height. One object falls straight down, one slides down a low-fric

tion inclined plane, and one swings in a circular arc on the end of a string. All three objects end at the same height.
In which case does the object have the biggest total work done on it by all forces during its motion?
A. Free Fall
B. Incline
C. String
D. Same
Physics
1 answer:
erik [133]2 years ago
4 0

Answer:

D. Same

Explanation:

Because only gravity is doing the work on the objects, and gravity is constant for all the objects

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Alik [6]

Answer:

The correct answer to the following question will be Option A (moment arm; pivot point).

Explanation:

  • The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
  • The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
  • The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.

The other three choices are not related to the given situation. So that option A is the appropriate choice.

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2 years ago
A yo-yo can be thought of as a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (s
lbvjy [14]

Answer:

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

now T= mg-ma

net torque acting on the yo-yo is

τ_net = Iα

I= moment of inertia (= 0.5 mr^2 )

α = angular acceleration

τ_net = 0.5mr^2(a/r)

Tr= 0.5mr^2(a/r)

(mg-ma)r=0.5mr^2(a/r)

a(1/2+1)=g

a= 2g/3

a= 2×9.8/3 = 6.5 m/s^2

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2 years ago
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melisa1 [442]

X = 3 · log(Y²)

X = 3 · 2·log(Y)

X/6 = log(Y)

10^(X/6) = 10^log(Y)

Y = 10^(X/6)

6 0
2 years ago
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Two 8.0 Ω lightbulbs are connected in a 12 V series circuit. What is the power of both glowing bulbs?
V125BC [204]

Answer:

18 W

Explanation:

Applying,

P = V²/R.................. Equation 1

Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs

Since: It is a series circuit,

Then,

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Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb

Given: R1 = R2 = 8 Ω

Substitute into equation 1

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R = 16 Ω

Also Given: V = 12 V

Substitute into equation 1

P = 12²/8

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P = 18 W

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2 years ago
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