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2 years ago

crowbar of 5 metre is used to lift an object of 800 metre if the effort arm is 200cm calculate the force applied

Physics

1 answer:

Klio2033 [76]2 years ago

4 0

Answer:

F = 5226.6 N

Explanation:

To solve a lever, the rotational equilibrium relation must be used.

We place the reference system on the fulcrum (pivot point) and assume that the positive direction is counterclockwise

F d₁ = W d₂

where F is the applied force, W is the weight to be lifted, d₁ and d₂ are the distances from the fulcrum.

In this case the length of the lever is L = 5m, t the distance desired by the fulcrum from the weight to be lifted is

d₂ = 200 cm = 2 m

therefore the distance to the applied force is

d₁ = L -d₂

d₁ = 5 -2

d₁= 3m

we clear from the equation

F = W d₂ / d₁

W = m g

F = m g d₂ / d₁

we calculate

F = 800 9.8 2/3

F = 5226.6 N

Read more on Electric Force:

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7 0

2 years ago

What relationship exists between speed, velocity, and acceleration

enyata [817]

Velocity is speed and the direction of the speed. Acceleration is the rate at which Velocity is changing, and the direction in which it's changing.

6 0

3 years ago

An orchestra is practicing a piece that is to be played at an allegro tempo. The conductor sets a metronome at 160 beats per min

omeli [17]

Answer:

375 ms

Explanation:

the frequency of metronome , f = 160 beats per minute

f = 160 /60 beats per sec

f = 2.67 beats /s

the period of a single beat , T = 1/f

T = 1/2.67 s

T = 0.375 s = 375 ms

the period of a single beat is 375 ms

3 0

3 years ago

HELP!!!!!! Need Help !!!

denis23 [38]

Answer:

its y

Explanation:

3 0

2 years ago

A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l

son4ous [18]

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

$\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Using formula of focal length for glass ethyl alcohol system

$\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Divided equation (II) by (I)

$\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}$

Where, $n_{g}$ = refractive index of glass

$n_{a}$ = refractive index of air

$n_{ethyl}$ = refractive index of ethyl

Put the value into the formula

$\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}$

$\dfrac{f'}{11.5}=\dfrac{25}{7}$

$f'=\dfrac{25}{7}\times11.5$

$f'=41.07\ cm$

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

7 0

3 years ago

crowbar of 5 metre is used to lift an object of 800 metre if the effort arm is 200cm calculate the force applied​

crowbar of 5 metre is used to lift an object of 800 metre if the effort arm is 200cm calculate the force applied