Question 5

If a person is near sighted what type of lens will their glass have to correct their eyesight

oksian1 [2.3K]

Answer:

Since the nearsighted eye over converges light rays, the correction for nearsightedness is to place a diverging spectacle lens in front of the eye. This reduces the power of an eye that is too powerful.

Explanation:

7 0

3 years ago

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I will mark as the brainliest answer plz 8,9,10

blagie [28]

Answer:

8. acceleration = $\dfrac{d(velocity)}{d(time)}$ = 1 unit .

9. acceleration = $\dfrac{d(velocity)}{d(time)}$ = -1 unit.

10. acceleration = $\dfrac{d(velocity)}{d(time)}$ = 0 units.

Explanation:

8. i) acceleration = velocity / time

ii) In this figure velocity = time

iii) therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = 1 unit .

9. i) acceleration = velocity / time

ii) In this figure 4 = m + 5, therefore m = -1

therefore velocity = (-0.5 $\times$ time) + 5

iii) therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = -1 units.

10.) velocity is constant at 2

therefore acceleration = $\dfrac{d(velocity)}{d(time)}$ = 0 units

5 0

4 years ago

A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m

madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂ = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

4 0

4 years ago

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy

Nady [450]

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

$E_i=\dfrac{mv^2}{2}+mgh$

Here , initial velocity is zero .

Therefore , $E_i=mgh=40\ J$

Now , final energy of the system :

$E_f=\dfrac{mv^2}{2}+mg(0)+15\\\\E_f=\dfrac{0.05\times v^2}{2}+15$

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

$E_i=E_f\\40=\dfrac{0.05v^2}{2}+15\\\\25=\dfrac{0.05v^2}{2}\\\\v=31.62\ m/s$

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

3 0

3 years ago

The nucleus of an atom has a positive charge. True False

xxTIMURxx [149]

The answer is True they have a positive charge

4 0

3 years ago

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