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zloy xaker [14]
3 years ago
12

The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f

rom the eye. Does the older person’s eye have a larger or smaller refractive power than a normal eye, when focused on an object at the near point? Explain.
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

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The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a
Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

with m the mass , a the acceleration and \sum\overrightarrow{F} the net force forces that is:

(F-f) (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

(F-f)=ma

solving for a:

a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}

Now let's use the Galileo’s kinematic equation

Vf^{2}=Vo^{2}+2a\varDelta x (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and \varDelta x the time took to achieve that velocity, solving (3) for \varDelta x:

\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}

t=68.94 m

8 0
3 years ago
A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how
Tamiku [17]

Answer:

W =  343.2 J

Explanation:

Given that,

Mass of bale of hay = 26 kg

Horizontal force exerted = 88 N

Distance moved, d = 3.9 m

Work done, W = Fd

Put all the values,

W = 88 N × 3.9 m

= 343.2 J

So, the work done is 343.2 J.  

7 0
3 years ago
A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A)
lord [1]

Answer:

4.53482 m/s

4.506 m/s

Explanation:

m_1 = Mass of player = 75 kg

v_1 = Initial velocity of player = 4.6 m/s

m_2 = Mass of ball = 0.47 kg

v_1 = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s

The player's speed is 4.506 m/s

4 0
4 years ago
Please help. Having a hard time figuring out
Goryan [66]
Yeah that’s is correct
5 0
3 years ago
CAN I HAVE SOME HELP PLEASE
AURORKA [14]

Answer:

By teh way is isn't it question of law or science and the picture is of what electric light or not I have read it so I was asking isn't is question of science

5 0
3 years ago
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