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EastWind [94]
3 years ago
6

A real image is four times as far from a lens as is theobject.

Physics
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

1.25 focal lengths

Explanation:

The lens equation states that:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the object distance

q is the image distance

In this problem, the image is 4 times as far from the lens as is the object: this means that

q=4p

If we substitute this into the lens equation and we rearrange it, we get

\frac{1}{f}=\frac{1}{p}+\frac{1}{4p}=\frac{4+1}{4p}=\frac{5}{4p}\\p=\frac{5}{4}f=1.25 f

so, the object distance measured in focal lengths is

1.25 focal lenghts

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Answer:

The acceleration increases.

Explanation:

From Newton's 2nd Law, we have \Sigma F=ma. We can see that force is directly proportional to mass and acceleration. Therefore, as force increases, either mass or acceleration must increase as well, and vice versa. Since mass is maintained here, if you increase the force applied to the Frisbee, the acceleration will increase as well.

8 0
3 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, 
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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>
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heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence, 
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a) </span>
Total height = 525+120.54 = 645.54 meters 

b) 
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4 0
3 years ago
Read 2 more answers
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Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

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