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natali 33 [55]
3 years ago
6

A farsighted boy has a near point at 2.3 m and requires eyeglasses to correct his vision. Corrective lenses are available in inc

rements in power of 0.25 diopters. The eyeglasses should have lenses of the lowest power for which the near point is no further than 25 cm. The correct choice of lens power for eyeglasses, in diopters, is:
Physics
1 answer:
tino4ka555 [31]3 years ago
5 0

Answer:

P = 3.5 D

Explanation:

As we know that convex lens is to be used to make the near point of eye to be correct

So we will have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we have

d_i = 2.3 m = 230 cm

d_o = 25 cm

now plug in all values into the formula

-\frac{1}{230} + \frac{1}{25} = \frac{1}{f}

f = 28 cm

now for power of lens

P = \frac{1}{f}

P = \frac{1}{0.28} = 3.5 D

so the power in dioptre is

P = 3.5 D

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2. A driver brings a car traveling at 22 m/s to a stop in 2.0 seconds. What is the car's acceleration?
sveta [45]

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11 m/s²

Explanation:

Acc = v/t

Acc = 22 / 2.0

Acc = 11 m/s²

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A slender, uniform metal rod of mass M and length lis pivoted without friction about an axis through its midpoint andperpendicul
lys-0071 [83]

Answer:

Explanation:

Moment of inertia of the metal rod pivoted in the middle

= M l² / 12

If the spring is compressed by small distance x twisting the rod by angle θ

restoring force by spring

= k x

moment of torque  about axis

= k x l /2

= k θ( l /2 )²     ( x / .5 l = θ )

=

moment of torque = moment of inertia of  rod  x angular acceleration

k θ( l /2 )²   = M l² / 12 d²θ/dt²

d²θ/dt² = 3 k/M  θ

acceleration =  ω² θ

ω² = 3 k/M

ω = √ 3 k / M

8 0
3 years ago
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Can someone help quickly please and thank you
Sergio039 [100]

Answer:

I think D??

Explanation:

8 0
3 years ago
A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the
Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0
3 years ago
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