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Lerok [7]
3 years ago
14

A farsighted girl has a near point at 2.0 m but has forgotten her glasses at home. The girl borrows eyeglasses that have a power

of 2.75 diopters. These are not going to make her able to view normally. With these eyeglasses, what is the new near point of the girl, assuming that she wears them extremely close to her eyes

Physics
1 answer:
lisov135 [29]3 years ago
6 0

Find the pictures in the attachment

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What is scientific evidence and when do we have enough?
Mandarinka [93]
Scientific evidence is evidence that is backed up by science and you have enough when you can understandingly prove your point or hypothesis
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2 years ago
Help! PROJECTILE PROBLEM: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25
dsp73
Here, 

height at failure, h1 = 525 m, 
upward acceleration, a = 2.25 m/s^2, 
velocity = v m/s, 
<span>
SO, </span>
<span>
v^2 = 2*a*h = 2*2.25*525 = 2362.5 </span>
Now, acceleration, g = 9.8 m/s^2, 
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SO, </span>
<span>
heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence, 
<span>
a) </span>
Total height = 525+120.54 = 645.54 meters 

b) 
<span>time, for h1, t = v/g = sqrt(2362.5)/9.8 = 4.96 sec 

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4 0
3 years ago
Read 2 more answers
Your physics textbook is sliding to the right across the table draw the vectors starting at the black dot. the location and orie
liberstina [14]
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6 0
3 years ago
Simplify -2/3 - 3/5<br><br> A) -1/15<br><br> B) -19/15<br><br> C) -3/5<br><br> D) -5/8
Leto [7]
I got B,when you subtract 3/5 from NEGATIVE 2/3 it creates a negative 19 over a positive 15.

7 0
3 years ago
A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t
Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

                                                    = 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0
3 years ago
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