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blagie [28]
2 years ago
13

Concave lenses have

Physics
1 answer:
Scilla [17]2 years ago
3 0

Answer:

A. negative; virtual.

Explanation:

A concave lens always forms a virtual erect and diminished image. The reason why is that the image is actually formed by the intersection of virtually extended refracted rays.

Hope this helps!

Please mark as brainliest if correct!

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. If block A has a velocity of 0.6 m/s to the right, determine the velocity of cylinder​
andrezito [222]

Answer:

As we can see, a string is attached with block A, and three string is folded with ply which is attached with B

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Given,

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​

Explanation:

IF THE ANSWER IS RIGHT PLZ GIVE ME BRAINLIEST

THANK U

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3 years ago
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Which of the following best describes the circuit shown below?
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A..............hope it will help you

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Which statement is true about Gram negative organisms?
aliina [53]

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B, C and D are true.

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3 years ago
The equation below can be used to find the specific heat capacity of a substance. What is the specific
slavikrds [6]

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6 0
2 years ago
Two simple pendulums are in two different places. The length of the second pendulum is 0.4 times the length of the first pendulu
faltersainse [42]

Answer:

\sqrt{\frac{4}{9}}

Explanation:

The frequency of a simple pendulum is given by:

f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}

where

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L is the length of the pendulum

Calling L_1 the length of the first pendulum and g_1 the acceleration of gravity at the location of the first pendulum, the frequency of the first pendulum is

f_1=\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}

The length of the second pendulum is 0.4 times the length of the first pendulum, so

L_2 = 0.4 L_1

while the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum, so

g_2 = 0.9 g_1

So the frequency of the second pendulum is

f_2=\frac{1}{2\pi}\sqrt{\frac{g_2}{L_2}}=\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}

Therefore the ratio between the two frequencies is

\frac{f_1}{f_2}=\frac{\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}}{\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}}=\sqrt{\frac{0.4}{0.9}}=\sqrt{\frac{4}{9}}

8 0
3 years ago
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