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Karo-lina-s [1.5K]
3 years ago
6

What are the characteristics of image when object is between f1 and 2f1 for concave lense?​

Physics
1 answer:
k0ka [10]3 years ago
5 0

Answer:

The required diagram is shown in the figure. When an object is placed in front of the convex lens, i.e., between 2F

1

and F

1

, its image is formed beyond 2F

2

on the other side of the lens. The image is real, inverted and enlarged.

solution

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bixtya [17]
Cutting pepper and onions
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3 years ago
A walkman uses four standard 1.5 V batteries. How much resistance is in the circuit if it uses a current of 0.02A? *
hodyreva [135]

Answer:

75ohms

Explanation:

V= IR

V = 1.5volts

I = 0.02A

1.5 = 0.02×R

Making R the subject

R = 1.5/0.02

R = 75ohms

The resistance in the circuit will be 75ohms

7 0
3 years ago
Food that is cooked properly can no longer be contained. True or false
Roman55 [17]

Answer:

true they can no longer be contained

4 0
3 years ago
What are two ways to create a negative ion?<br> What are two ways to create a positive ion?
Iteru [2.4K]

Answer:

what that guy said

Explanation:

because he provides evidence

4 0
3 years ago
A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00 m above the ground. At th
Elis [28]

Answer:

  • <u>The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.</u>

Explanation:

The motions of the two water ballons are ruled by the kinematic equations:

  • y=y_0+V_0t-gt^2/2

We are only interested in the vertical motion, so that equation is all what you need.

<u>1. Water ballon is thrown horizontally at sped 2.00 m/s.</u>

The time the ballon takes to hit the ground is independent of the horizontal speed.

Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.

Then:

y=y_0+V_0t-gt^2/2\\ \\ 0=6.00m-9.8\frac{m}{s^2} t^2/2\\ \\ t=\sqrt{2\times6.00m/9.8\frac{m}{s^2}}\\\\ t=1.11s

<u>2. Water ballon thrown straight down at 2.00 m/s</u>

Now the initial vertical speed is 2.00 m/s down. So, the equation is:

0=6.00m-2.00\frac{m}{s}t-9.8\frac{m}{s^2}t^2/2\\ \\ 4.9t^2+2t-6=0\\ \\ t=0.92s

To solve the equation you can use the quadratic formula.

t=\frac{-2+/-\sqrt{2^2-4(4.9)(-6)} }{2(4.9)}\\ \\ t=-1.33\\ \\ t=0.92

You get two times. One of the times is negative, thus it does not have physical meaning.

<u>3. Conclusion:</u>

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

5 0
3 years ago
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