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blagie [28]
2 years ago
15

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. A person can clearly see objects only if they are located between 45 cm and 161 cm away from her eyes. Bifocal lenses are used to correct her vision. What power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?
Physics
1 answer:
Anarel [89]2 years ago
4 0

Answer:

P= -0.62 D

Explanation:

We know that lens equation

1/f=1/u+1/v

Given that person can see object only when object is located between 45 and 161 cm so we can say that far point of that person is at 161 cm.

So we need to take that objects ray are coming from and infinity and will focus at 161 cm away.

So now by putting the values

1/f=1/∞+1/-161    (negative because image will be virtual ,by sign convention)

So f=-161 cm

We know that inverse of focus(in meter) length is called power of lens.

So power of lens P=1/f

  P=-1/1.61

P= -0.62 D

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lord [1]

Answer:

To both observers, the land opposite them is moving to the right.

Explanation:

I have this class too. and there is also a quizlet with all the answers to the rest of the other questions. Trust me its right .

https://quizlet.com/261219090/oce-1001-chapter-2-flash-cards/

5 0
2 years ago
before colliding, the momentum of block A is +15.0 kg m/s. after, block A has a momentum -12.0 kg*m/s. what is the momentum of b
Helen [10]

Answer:

The momentum of block B = 27 Kg m/s

Explanation:

Given,

The initial momentum of block A, MU = 15 Kg m/s

The final momentum of block A, MV = -12 Kg m/s

Consider the block B is initially at rest.

Therefore, the initial momentum of block B, mu = 0

According to the laws of conservation of linear momentum, the momentum of the body before impact is equal to the momentum of the body after impact.

                               <em> MU + mu = MV + mv</em>

                                15  +  (0) = (-12) + mv

                                         mv = 15 + 12

                                              =  27 Kg m/s

Hence, the momentum of the block B after impact is, mv = 27 Kg m/s

3 0
2 years ago
Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom
Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

B = B_o(-\hat k)

now the velocity is from bottom to top

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v = v_o \hat j

now the force on the moving charge is given as

\vec F = q(\vec v \times \vec B)

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\vec F = (-e)(v_o \hat j \times B_o(-\hat k))

\vec F = e v_o B \hat i

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7 0
2 years ago
A toy plane has a mass of 2.5 kg and is 18 m above the ground. It is moving at 4.5 m/s. How much mechanical energy does the toy
sleet_krkn [62]
Mechanical energy is the sum of kinetic energy and potential energy, or E=Ek+Ep. So Ek=(1/2)*m*v² where m is the mass of the object and v is it's velocity. Mp=m*g*h where m is the mass, g=9.81 m/s² and h is the height of the object. So after we input the numbers the total mechanical energy is
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4 0
2 years ago
Read 2 more answers
A 1.25 in. by 3 in. rectangular steel bar is used as a diagonal tension member in a bridge truss. the diagonal member is 20 ft l
pentagon [3]

Answer:

axial stress in the diagonal bar =36,000 psi

Explanation:

Assuming we have to find axial stress

Given:

width of steel bar: 1.25 in.

height of the steel bar: 3 in

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modulus of elasticity E= 30,000,000 psi

strain in the diagonal member ε = 0.001200 in/in

Therefore, axial stress in the diagonal bar σ = E×ε

=  30,000,000 psi×  0.001200 in/in =36,000 psi

5 0
2 years ago
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