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alekssr [168]
2 years ago
10

What type of lens is used to correct nearsightedness.

Physics
1 answer:
Stella [2.4K]2 years ago
5 0

Answer:

Concave lenses are used in eyeglasses that correct myopia or nearsightedness.

You might be interested in
A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute
levacccp [35]

Explanation:

The given data is as follows.

          Mass, m = 75 g

         Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

          m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

where,  m_{1} = mass of the projectile

            m_{2} = mass of block

              v = velocity after the impact

Now, putting the given values into the above formula as follows.

              m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

         75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v

                                  = \frac{45}{50.075}

                              v = 0.898 m/s

Now, equation for energy is as follows.

               E = \frac{1}{2}mv^{2}

                  = \frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}

                  = 13500 J

Now, energy after the impact will be as follows.

             E' = \frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}

                 = 20.19 J

Therefore, energy lost will be calculated as follows.

           \Delta E = E  E'

                       = (13500 - 20) J

                       = 13480 J

And,   n = \frac{\Delta E}{E}

             = \frac{13480}{13500} \times 100

             = 99.85

             = 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0
3 years ago
Read 2 more answers
Which of Jafar's statements are correct regarding distance and displacement?
shepuryov [24]

Answer:The distance and magnitude of displacement are sometimes equal." Jafar is correct. The distance traveled and the magnitude of displacement are equal if and only if the path is a straight line in one direction.

Explanation:

5 0
3 years ago
Which of the following is a likely life cycle of a star?
melomori [17]

Answer:

D.

Explanation:

But this just happen for big stars, like more than 20x the Sun mass.

Shortly: A nebula is a cloud of gas and dust, the material starts to be acummuleted and became a protostar (is like a big planet, almost  a star). With enought mass this is a star, burn hydrogen and transform it in Helium.

This occurs in Main Sequence, is about almost all the life time of a star. Then starts the lack of hydrogen. Gravity compress everything, pressure goes up and heat all. Too much energy, Helium get burned and the star grews fast, became a Red Giant. Time pass and the fuel is over, no more making fusion, gravity compress the star, too much strenght, colapses, neutron star.

If it have pretty mass, ok. If have more than like 2x Sun mass, became a blackhole.

7 0
3 years ago
Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri
igor_vitrenko [27]

Answer:

a) the particles are <em>0.217 m </em>apart

b) <em>the particles are moving in the same direction</em>.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

    = -0.155 A + 0.372 A

    = 0.217 A

since A = 1 m

Thus,

<em>Δx  = 0.217 m</em>

<em></em>

<em></em>

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

   = (-πA / T) sin(2πt / T)

   = -(π(1) / 1.5) sin(2π(0.45) / 1.5)

   = -1.99

and,

v₂ = dx₂ / dt

   = (-πA / T) sin((2πt / T) + π/6)

   = -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

   = -1.40

Since both v₁ and v₂ are negative, this shows that <em>the particles are moving in the same direction</em>.

6 0
3 years ago
A wood block, after being given a starting push, slides down a ramp at a constant speed. what is the angle of the ramp above hor
Stells [14]

The solution for the problem is:

Constant speed means Fnet = 0. 
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …

The computation would be:


Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ) 
mg (sin Θ) = µk (mg) (cos Θ) 
µk = tan Θ 
Θ = arctan(µk)

= arctan (0.35)

≈ 19.3°

5 0
3 years ago
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