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Damm [24]
3 years ago
6

The near point of an eye is 56.0 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan

ce of the healthy near point. What should be the focal length of this lens
Physics
1 answer:
irakobra [83]3 years ago
4 0

Answer:

Explanation:

Near point = 56 cm .

near point of healthy person = 25 cm

person suffers from long sightedness

convex lens will be required .

object distance u  = 25 cm

image distance   v = 56 cm

both will be negative as both are in front of the lens.

lens formula

I/v - 1 / u = 1/f

- 1/56 +1/25 = 1/f

- .01785 + .04 = 1/f

1/f  = .02215

f = 45.15 cm .

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MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

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7 0
2 years ago
A rocket initially at rest accelerates at a rate of 99. 0 meters/second2. Calculate the distance covered by the rocket if it att
creativ13 [48]

The rocket will cover 1 \times  10^3\rm \ m distance in 4. 5 s. Acceleration can be defined as the change in velocity.

<h3>What is acceleration?</h3>

Acceleration can be defined as the change in speed or the direction of the object.

From kinamatic equation:

D = v_{t} +\dfrac 12at^2

Where,

D - final velocity =  445 m/s

v_0 -  initial valocity = 0 m/s

a - acceleration = 99. 0 m/s²

t - time =  4. 50 s

Put the values in the formula,

D = 0\times  ( 4.5) + \dfrac12\times (99)(4.5)^2\\\\D = 1002 {\rm \ m}\\\\D = 1 \times  10^3\rm \ m

Therefore, the rocket will cover 1 \times  10^3\rm \ m distance in 4. 5 s.

Learn more about Acceleration :

brainly.com/question/2697545

7 0
2 years ago
A uniform horizontal beam 4.0 m long and weighing 200 N is attached to the wall by a pin connection that allows the beam to rota
wariber [46]

Answer:

The magnitude of the tension in the cable, T is 1,064.315 N

Explanation:

Here we have

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Weight = 200 N

Center of mass of uniform beam = mid-span = 2.0 m

Point of attachment of cable = Beam end = 4.0 m

Angle of cable = 53° with the horizontal

Tension in cable = T

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Weight of person = 350 N

Therefore,

Taking moment about the wall, we have

∑Clockwise moments = ∑Anticlockwise moments

T×sin(53) = 350×1.5 + 200×2

T = 850/sin(53)  = ‭1,064.315 N.

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